Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 1

Latent Heat, Phase Change

Phase Change: Basic concept, definitions and facts

Matter normally exists in one of the three phases: solid, liquid and gas1 . Solid is melted into liquid; liquid is boiled or evaporated into gas as the heat is added on to the system. Conversely, gas condenses into liquid and liquid freezes into solid if the heat is taken out of the system by some means. The heat that is required to be added to a system or that is given up by the system for any kind of phase change to occur is the usual sensible heat as discussed in the previous chapter. However, during the time of phase change, the temperature of a system remains unchanged, although the heat is being transferred to or from the substance. The thermometer shows no increase (or decrease) of temperature. This is the concept of latent heat which is a ‘hidden’ heat. Sensible heat turns to latent heat during a phase change.

Where does the heat go?

Addition of heat causes increased vibration of molecules about their equilibrium positions in a solid and the motion of molecules in a liquid. As a result of this, the temperature of a substance increases. During a phase change (or phase transition), the heat energy goes into the work of breaking bonds or separating molecules etc. rather than increasing the temperature of a substance.

Where do we commonly observe phase change?

Melting of ice and boiling of water are common things. Ice starts melting at 0 0 C . The temperature remains fixed at 0 0 C until the entire ice melts and is converted to water. When the phase change from solid ice to liquid water is over, the temperature of water is seen to rise further if heat is being added. Again as the temperature of water reaches 100 0 C , the liquid water starts boiling and another phase change process starts. The temperature of boiling water remains fixed at 100 0 C during this time until the entire water is converted to vapour. If now more heat is added, the vapour/ steam can only be superheated (temperature becomes higher than 100 0 C ).

1 Plasma is often considered a fourth state of matter. Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 2

d e 100 Temp. c ( 0 C ) b 0 a -50 Time

The above figure demonstrates how the temperature changes with time during phase transitions (ice → water → vapour): a. Temperature rises up to 00 C as the ice absorbs heat. b. Latent heat is absorbed by ice and it starts melting (Temp. remains fixed) c. Temperature rises again as the liquid water absorbs heat. d. Water absorbs latent heat of vaporization and starts boiling at1000 C (Temp. remains fixed) e. Heat is absorbed by steam and thus the temperature increases again.

Table: 1

Phase Changes of various types: • Melting/ Fusion: Solid Liquid • Vaporization: Liquid Vapour (gas) 1. Evaporation (slow) 2. Boiling (fast) • Sublimation: Solid Vapour • Condensation: Vapour Liquid • Freezing/ Solidification: Liquid Solid • Deposition: Vapour Solid

Concept of Critical Temperature:

For every pure substance, there is a critical temperature (corresponding to a fixed pressure) at which the process of phase change begins. This is termed as critical point. The critical temperature at which the ice starts melting is 0 0 C (at normal atmospheric pressure) and this is called melting point of ice. The critical temperature at which water starts boiling is 100 0 C (at normal atmospheric pressure) and this is called boiling point Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 3 of water. Similarly, there is freezing point at a temperature when freezing (or solidification) process starts and so on. For every pure substance and for every kind of phase change there is a definite critical point (the value of which depends on the atmospheric pressure). However, the impure or mixed substances have no definite melting or freezing point. Solids of different kinds are mixed together in different proportions to prepare mixtures whose melting points are estimated for practical and experimental purposes. Similarly, different liquids can be mixed to have solutions of varied freezing points.

Change of volume during phase change:

Because of melting the solid becomes liquid and due to freezing the liquid becomes solid again. In most of the cases, the solids have higher densities and smaller volumes than their corresponding liquid states. There are some exceptions. We observe that a piece of ice floats on water whereas a piece of solid wax sinks. The solid ice has lower density (and thus greater volume) than the water. Some other materials like cast iron, brass, antimony, bismuth etc. show similar behaviour like that of ice. Good castings (making dice) can be made using the property of the expansion of solid, like cast iron, which expands due to solidification. When the water freezes in winter, in cold countries, there is often enormous pressure exerted by ice in the water pipes. The bursting of pipes and the cracking of rocks in the mountains are often due to this.

Effect of Pressure on Melting point:

• For the materials which expand due to melting, the melting points of those materials become higher due to increase in pressure on them. This means, the materials now melt at higher temperatures due to increased pressure on them.

• The materials which contract due to melting (or expand due to freezing), such as ice, the melting points of them come down due to increase in external pressure. This means, the materials now melt at lower temperatures due to increased pressure on them. For example, a pressure of 500 atm forces the ice to melt at − 40 C .

We have often observed two pieces of ice can be joined by pressing them together. The pressure exerted upon them makes the melting point lower where two faces join (interface). The ice melts at a temperature lower than 0 0 C due to excess pressure and when the pressure is withdrawn, the melted water at that site freezes again which makes the two ice pieces unite. This phenomenon is called regelation. Regelation was discovered by Michael Faraday.

Regelation: This is the phenomenon in which water refreezes to ice after it has been melted by pressure at a temperature below the freezing point of water.

Examples: Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 4

• Pressure makes an ice skate, form a film of water that freezes once again after the skater has passed.

• A weighted wire slowly melts through a block of ice whereas the ice refreezes behind it. This stops the ice block from breaking in half. This is known as Bottomley’s experiment.

What is a freezing mixture?

Freezing mixture is a mixture of substances whose freezing point is lower than that of its constituents.

A commonly used freezing mixture is ice with salt (NaCl). For example, the freezing point of a 1 M NaCl solution is − 3.4 °C, approximately. Any solutions will have such a lowering or depression of freezing point. In case of saline water, the higher the concentration of salt, the greater is the freezing point depression. Any foreign substance added to the water will cause a freezing point depression.

In cold countries, the water in lake or river freezes in winter. However, the sea water does not easily freeze. The freezing point of saline water in sea comes down below 0 0 C .

Note: For example, normal seawater, which contains approximately 3500 parts per million salt 0 (including Na, Ca, Cl, Mg, K, SO 4 etc.), freezes at around − 2.2 C . In case of water with extreme salinity, such as very salty lake waters at Death Valley, California (approximately 300,000 parts per million salt content) may freeze at a temperature as low as − 200 C .

Ice mixed with salt melts quickly. Why does salt melts ice?

The following is a microscopic explanation: Ice starts melting at 0 0 C and melts at any temperature above this. Two opposite things happen at this time. Molecules on the surface of ice escape into water (melting) and the molecules of water are captured on the surface of ice (freezing). The water and ice are said to be in dynamic equilibrium in this way. Adding salt to the system disrupts this equilibrium. Salt gets easily dissolved into water. Now some of the water molecules are replaced by salt molecules. So, the number of water molecules captured by ice per second goes down and thus the rate of freezing goes down. On the other hand, the rate of melting is unchanged by the presence of the foreign molecules (salt). So, so melting occurs faster than freezing. This makes the ice melt quicker.

Vaporization by Evaporation and Boiling:

Vaporization is a phase change process, in general, whereby a liquid changes to gaseous phase. This may happen in the following ways: Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 5

• Evaporation: Evaporation is the process of vaporization whereby atoms or molecules from the surface of a liquid phase gain sufficient energy to enter the gaseous phase. This is the opposite process of condensation. (In case of molecules or atoms evaporating from a solid surface, it is known as sublimation. Example is camphor.) The process of evaporation is an important component of the hydrological cycle in nature. Solar energy drives evaporation of water from oceans, lakes, rivers, moisture in soil etc. to form clouds and rain.

• Boiling: When vaporization takes place from all parts of the liquid, as a critical temperature is reached (boiling point), the phenomenon is called boiling.

Microscopic Explanation of Boiling:

• As a liquid is heated, its vapour pressure increases until the vapour pressure becomes equal to the pressure of the gas above it. • Bubbles of the vaporized liquid (i.e., gas) form within the volume of the liquid and then rise to the surface where they burst and release the gas. At the boiling point, the vapour inside a bubble has enough pressure to keep the bubble from collapsing. • In order to form vapour, the molecules of the liquid must overcome the forces of attraction between them.

What is the effect of pressure on boiling point?

The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere (76 cm Hg). When the external pressure is less than one atmosphere, the boiling point of the liquid is lower than its normal boiling point. When the external pressure is greater than one atmosphere, the boiling point of the liquid is greater than its normal boiling point.

At high altitude, like on a hill resort or on a high mountain, because of lower atmospheric pressure, water boils at a temperature much below its normal boiling point which is 100 0 C . Thus there is a difficulty in cooking food properly at high altitudes. Pressure cooker is used to overcome the difficulty. The pressure cooker is a sealed vessel which does not allow the vapour to escape below a preset pressure. The pressure is developed above the liquid surface inside the vessel. As the boiling point of a liquid increases as the pressure increases, the liquid is made to boil a t a higher pressure. Water boils at 120 0 C in a pressure cooker. The higher temperature causes the food to cook faster. Cooking times can be greatly reduced (Almost 70 percent time and accordingly fuel can be saved).

Fig. to be included Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 6

Triple Point of water:

For all substances, as we lower pressure, the boiling temperature falls much more rapidly than does the freezing temperature. For water, the freezing temperature rises slightly at low pressure. Thus we can ask the question: Are the boiling temperature and freezing temperature ever equal? The answer is yes.

At the low pressure of around 0.006 atm ( = 611 Pa), pure water boils at 0.01 0 C , and it also freezes at 0.01 0 C . The combination of conditions (P, T) = (611 Pa, 0.01 0 C ) is called the triple point of water because, at this pressure and temperature ice, liquid water and steam can coexist in equilibrium. This point is used to define the scale of temperature.

How can we determine altitude from boiling point?

Boiling point decreases due to decrease in atmospheric pressure on high altitude. It is seen that the boiling point of water comes down by 1 0 C as the atmospheric pressure is decreased by 27 cm Hg, on average. On the top of Mount Everest (8848 m), the water boils approximately at 70 0 C . If the height of a mountain is H cm from the sea level and the pressure difference between the sea level and the top of the mountain is h cm, we can write,

The pressure by H cm of air column = the pressure by h cm of mercury column Or, H × ρ × g = h ×13.6× g [ ρ = the average density of air at that place] 13.6 ∴ H = h. ρ The above simplified relation is however, not true. In reality, density ρ is not a constant; it varies greatly with air pressure.

Other factors affecting boiling point:

• Salt, sugar and practically any other substance help increasing the boiling point. Cooking time can be shortened when these things are mixed with water. Water with high level of dissolved mineral salts is called hard water. Hard water boils at a higher temperature. • Boiling point goes down on a stormy or windy day.

The factors that affect evaporation:

1. Temperature: The higher the temperature, the higher the rate of evaporation. Evaporation takes place at all temperatures. But when the temperature of the liquid is increased the molecules at the surface gain more kinetic energy and leave the liquid surface at a faster rate.

Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 7

2. Wind: The stronger the wind, the higher the rate of evaporation. When evaporation takes place, liquid vapour gathers above the surface and wind helps it to remove the vapour as soon as it is formed. This helps further evaporation making more molecules to escape the surface.

3. Surface area: Evaporation increased as the exposed surface area of the liquid is increased. This is because the molecules escape only from the surface.

4. Humidity: If humidity is high, the rate of evaporation will be low. Humidity refers to the amount of water vapour (moisture) in the air.

Basic differences between Evaporation and Boiling:

1. Evaporation occurs at the surface of a liquid and the molecules escape from the surface. Boiling occurs over the entire volume of the liquid, bubbles are formed and molecules escape from all parts of the boiling liquid.

2. For evaporation of a liquid, the required latent heat is taken from the liquid itself (and also from surroundings). This results in cooling of the liquid. But in case of boiling, latent heat is taken from the supplied heat.

3. Evaporation occurs all the time and at all temperatures. In fact, temperature of a liquid is always changing during evaporation as the latent heat is continuously absorbed from the liquid itself. Boiling starts at a characteristic critical temperature (decided by the pressure) and the temperature remains fixed as long as the boiling continues.

4. Evaporation occurs when the vapour pressure is lower than the pressure of the gas above the liquid surface (the atmospheric pressure). Boiling of a liquid occurs when the vapour pressure is equal to the pressure of the gas above it.

5. Evaporation is slow and boiling is a fast process.

Sublimation:

Under appropriate conditions of pressure and temperature a solid substance can change to gas directly without going through the liquid phase. The transformation of solid to vapour is called sublimation. Example: camphor. Note: Solid carbon dioxide is called ‘dry ice’. Liquid carbon dioxide does not exist at a pressure more than 5 atm.

Latent Heat Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 8

Definition: The amount of heat that must be transformed when a unit mass of a substance undergoes a complete phase change is called latent heat.

[Note: Actually, the above should be the definition of specific latent heat. However, this is commonly referred as latent heat.]

Suppose, Q be the amount of heat involved in a phase change for a certain substance of mass m , we can write, Q = m.L . Here L is the latent heat.

When we consider the melting of a solid or boiling of a liquid, heat is added to the system and Q is taken as positive. In the case of condensation of vapor or freezing of a liquid, heat is released by the system and Q is taken as negative.

The latent heat (of fusion or melting) of ice at 0 0 C is 334 kJ/kg. This means that to convert 1 kg of ice at 0 0 C to 1 kg of water at 0 0 C , 334 kJ of heat must be absorbed by the ice. Conversely, when 1 kg of water at 0 0 C freezes to give 1 kg of ice at 0 0 C , 334 kJ of heat is released to the surroundings.

The units of L: In C.G.S. system, it is Cal/gm; in F.P.S. system, it is B.Th.U./ pound and in M.K.S. (or SI) system the unit is J/kg.

3 Note: 1 kJ = 10 J ; 1 Calorie = 4.2 J. So the units can be converted to KJ/kg or J/gm etc. as they are also used.

Table: 2

Substance Latent heat Melting Latent heat of Boiling of fusion point vaporization point (KJ/kg) ( 0 C ) (KJ/kg) ( 0 C ) Hydrogen 59 -259 452 -253 Oxygen 14 -219 213 -183 Nitrogen 26 -210 201 -196 Ethanol 109 -114 838 78 Mercury 12 -39 272 357 Ice/ Water 334 0 2258 100 Sulphur 54 115 1406 445 Silver 88 961 2193 2336 Gold 65 1063 1578 2660 Copper 134 1083 5069 1187 Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 9

1 KJ/kg = 1 J/gm = 1/4.2 cal/gm.

Latent heat of various kinds: Latent heat is different for different materials and for different types of phase transitions. Thus we come across various kinds of latent heat like latent heat of fusion for solid-to- liquid phase change; latent heat of solidification for liquid-to-solid phase change; latent heat of vaporization for liquid-to-gas phase change etc.

• Latent heat of fusion (melting): The heat energy required to change unit mass of a substance from the solid to the liquid phase without changing its temperature is called the latent heat of fusion of the substance.

• Latent heat of vaporization: The heat energy required to change unit mass of a liquid into the gaseous phase at the boiling point is called the latent heat of vaporization of the substance. For evaporation, there is no definite latent heat; it depends on the temperature of the liquid and other things.

• Latent heat of sublimation: Heat is absorbed in the process of sublimation. The quantity of heat absorbed per unit mass during sublimation is called the heat of sublimation.

Measurement of Latent heat of Vaporization of water:

Fig. to be included

Latent heat of steam can be measured by this method. Steam is produced by boiling of water in a flask B. The emerging steam from this is sent to a steam trap S through a bent tube L as shown. Two tubes C and D are inserted into the chamber S at the bottom. The Some steam is condensed into water and this drains out through the outlet C. The end of D inside S is bent. The dry steam jet comes out through the pipe D and enters into a calorimeter M, partly filled with water. There are a thermometer and a stirrer inside the calorimeter. The calorimeter is thermally insulated to resist heat loss.

Working principle: At the beginning, the empty calorimeter with the stirrer is weighed. Next, around half the calorimeter is filled with water and weighed again. Thus one obtains the mass of water in the calorimeter. The initial temperature of the water and the thermometer is measured by a thermometer T. Initially, some steam is allowed to pass through the outlet D so that after some time the dry steam jet comes out. The temperature of the steam can also be noted by a thermometer at this time. Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 10

Now the calorimeter with partly filled water is placed below the steam trap and the end of the tube D is immersed into the water. The steam condenses inside the cold water in the calorimeter. The calorimeter water is continuously stirred during the experiment. After some time, when the temperature of the calorimeter water increases to around 2-5 0 C , the calorimeter is removed. Now the calorimeter along with the warm water is weighed again and from this the initial weight of the calorimeter with water is subtracted. This gives the mass of the condensed steam.

Calculation: Let the latent heat of vaporization of water is L cal/gm, the water equivalent of the calorimeter is W gm, the mass of water in calorimeter is m1 gm and the mass of the condensed steam is m2 gm. Also let us assume that the initial temperature of the 0 calorimeter and water is t1 C and the final temperature of the calorimeter with water 0 and the condensed steam is t2 C . 0 0 Now, m2 gm steam at 100 C rejects latent heat and turns into m2 gm water at 100 C 0 and then this water further rejects heat to become m2 gm water at t2 C .

∴The total rejected heat = m2 L + m2 (100 − t2 ) cal. The above rejected heat is absorbed by the water in the calorimeter and the calorimeter itself. The total absorbed heat = W (t2 − t1 ) + m1 (t2 − t1 ) = (W + m1 )(t2 − t1 ) . ∴We can write,

m2 L + m2 (100 − t2 ) = (W + m1 )(t2 − t1 )

(W + m1 )(t2 − t1 ) Or, L = − (100 − t2 ) . m2 This relation can be used to determine the latent heat of vaporization of water.

In this experiment, the heat loss by radiation can be minimized by taking the initial temperature of the calorimeter water to be around the room temperature. The initial temperature of the calorimeter water may be taken to be around 2 0 C below the room temperature and the final temperature to be around 2 0 C above the room temperature.

Note: In the above, we have assumed the temperature of steam to be 100 0 C (at standard pressure). But it can actually be measured by a thermometer during the experiment and that value may be put in place of 100. Also, if the weight of the empty calorimeter (with stirrer) is M gm and the specific heat of the material of the calorimeter is S then we may use W = M.S in the formula.

Latent heat in calorimetric calculations:

In the calorimetric calculations (like those described in the previous chapter), where phase change occurs, we have to additionally consider latent heat.

Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 11

Let us suppose, an ice block of mass M gm at 0 0 C is plunged into the water of a calorimeter. Ice melts and is converted into water. Let the mass of water in calorimeter is m gm and the water equivalent of the calorimeter container is W gm. If the initial 0 0 temperature of calorimeter is t1 C and the final temperature is t2 C (when all the ice melts), we can write from the principle of calorimetry,

M.L + M (t2 − 0) = (W + m)(t2 − t1 ) .

Here, L cal/gm is the latent heat of melting ice. So the total absorbed heat consists of two parts: the latent heat that ice takes to melt and the heat that water from melted ice takes to reach the final equilibrium temperature. The total rejected heat comes from the calorimeter and water in it.

We have, M.(L + t2 ) = (W + m)(t2 − t1 ) (W + m)(t − t ) ∴ L = 2 1 − t M 2

Problems with Solutions

Note: • The problems presented here are similar to those given in the previous chapter on Calorimetry and Specific heat. One additional concept necessary is of the ‘latent heat’. • Most of the problems here are provided in C.G.S. system where the unit of heat is calorie (cal). Conversion to Joule (J) or Kilo Joule (KJ) can be easily done if we remember 1 cal = 4.2 J and 1 KJ = 1000 J.

Example 1: A 300 gm copper calorimeter contains 270 gm water. The temperature of the system is 30 0 C . Now 20 gm ice at −10 0 C is dropped in it. What will be the final temperature of the system? Assume the specific heat of copper = 0.1; specific heat of ice = 0.5; the latent heat of ice for melting = 80 cal/gm. Solution: Let the final temperature betC0 . This temperature must be somewhat between −10 0 C and 30 0 C . ∴The total heat rejected by the calorimeter and water = 300× 0.1× (30 − t) + 270×1× (30 − t) = (30 + 270).(30 − t ) = 300.(30 − t ) cal. (1) The heat absorbed has three parts: (i) Heat absorbed by ice at −10 0 C to reach at 0 0 C = 20× 0.5×10 = 100 cal. (ii) Latent heat absorbed by ice at 0 0 C to melt completely = 20 ×80 = 1600 cal. (iii) Heat absorbed by water from melted ice to reach the final temperature = 20 × t cal.

Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 12

∴The total heat absorbed in the system = 100 +1600 + 20.t = (1700 + 20.t ) cal. (2) Now, we can check if all the ice really melts. We put t = 0 and see that the amount of heat rejected as given by (1) is greater than the amount of heat absorbed as given by (2).

According to calorimetric principle, Total heat rejected = total heat absorbed. ∴ 300.(30 − t) = 1700 + 20.t Or, 9000 − 300.t = 1700 + 20.t Or, 320.t = 9000 −1700 7300 Or, t = = 22.81 320 ∴The final temperature of the mixture in the calorimeter is = 22.81 0 C

0 Example 2: Take a piece of ice at 0 C whose density is 0.916 gm/cc and the density of water is 1 gm/cc. Now a 10 gm metal piece is heated up to 100 0 C and then dropped into a system of ice and water. Some ice melts because of this and the volume of the mixture decreases by 0.1 cc whereas the temperature does not change. Find the specific heat of the metal if the latent heat for melting of ice is 80 cal/gm. Solution: We have to find out how much ice melts as we are given that the volume of ice-water mixture reduces by 0.1 cc. 1 The volume of 1 gm of ice at 0 0 C = = 1.092 cc. 0.916 1 The volume of 1 gm of water at 0 0 C = = 1 cc. 1 ∴When 1 gm ice melts, the reduction in volume = 1.092 −1 = 0.092 cc. 0.1 Now the reduction in volume by 0.1 cc is caused by the melting of = 1.087 gm 0.092 ice. The heat absorbed by 1.087 gm of ice for melting = 1.087 ×80 cal. This heat must be supplied by the heated metal piece. If S be the specific heat of the metal, the rejected heat by the metal = 10× S ×100 cal. ∴ 10 × S ×100 = 1.087 × 80 1.087 ×80 Or, S = = 0.087 1000

0 Example 3: The temperature of a 50 gm solid substance increases by 11 C in the first 1 minute as heat is supplied to it at the rate of 5 cal/sec. The temperature remains fixed for next 13 minutes. After this, the temperature starts increasing at the rate of 6 0 C /minute. Find (i) the specific heat of the substance when in solid phase, (ii) the specific heat when it becomes liquid, and (iii) the latent heat of melting. Solution:

(i) If the specific heat of the solid substance is S1 , we can write

50× S1 ×11 = 5× 60 Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 13

5× 60 ∴ S = = 0.545 1 50×11

(ii) As the temperature of the substance remains fixed at this second stage, the phase change happens which means the substance turns liquid from the solid form. If the latent heat of melting is L cal/gm, we can write 50× L = 13× 60× 5 Or, L = 78 ∴The latent heat of melting = 78 cal/gm.

(iii) At this stage, the temperature of the substance in liquid form increases as the heat is supplied.

If the specific heat of the substance in the liquid form is S 2 , we can write

50× S 2 × 6 = 5× 60 ∴ S 2 = 1.

Example 4: Calculate the total heat that must be supplied in order to convert 1 gm of ice at −10 0 C to steam of at 110 0 C at atmospheric pressure. Given, the specific heat of ice = 0.5 cal/(gm. 0 C ), the latent heat of melting of ice = 80 cal/gm, the latent heat of boiling of water = 540 cal/gm and the specific heat of steam at 1 atm = 0.48 cal/(gm. 0 C ). Solution: The total heat, required by the system, consists of the following five parts: (i) The heat absorbed by ice to raise its temperature from −10 0 C to 0 0 C = 1× 0.5×10 = 5 cal. (ii) The latent heat for melting of ice at 0 0 C = 1×80 = 80 cal. (iii) The heat required for the heating of water from 0 0 C to 100 0 C = 1× (100 − 0 ) = 100 cal. (iv) The latent heat for boiling of water at 100 0 C = 1× 540 = 540 cal. (v) The heat required for heating of vapour from 100 0 C to 110 0 C = 1× 0.48× (110 −100) = 4.8 cal. Thus the total heat that must be supplied to the system = (5 + 80 +100 + 540 + 4.8) = 729.8 cal.

Example 5: How much heat has to be removed by a refrigerator to convert 200 gm of water at 40 0 C to convert it to ice cube at − 4 0 C ? What will be the rate of work done by the refrigerator, if this formation of ice cube takes 2 minutes? Solution: At the first stage, the heat that is to be extracted to bring down the temperature of the water from 40 0 C to 0 0 C = 200×1× (40 − 0 ) = 8000 cal. At the second stage, the latent heat that is to be removed to convert water at 0 0 C to ice cube at the same temperature = 200×80 = 16000 cal. [ L = 80 cal/gm is the latent heat of fusion of ice.] Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 14

In the third stage, the refrigerator removes more heat to bring the temperature of the ice cube from 0 0 C down to − 4 0 C . The heat that has to be removed at this stage = 200× 0.5×[0 − (−4)] = 400 cal. [The specific heat of ice = 0.5] ∴The total heat that must be removed by the refrigerator = (8000 +16000 + 400) = 34400 cal = 34400 × 4.2 J = 144480 J = 144.48 kJ. 144.48 The rate of work done = kJ/sec. = 1.204 kW. 2× 60

[Note: In some books, the heat removed from a substance (or rejected by a substance) is indicated by a negative sign.]

0 Example 6: The tank on the roof-top of your house contains water at 10 C . Water from this tank enters in the geyser (water heater) in the bathroom where it is heated up to 90 0 C and then this hot water is drawn from the attached tap at the rate of 600 gm/min. What is the average electric power that is consumed by this geyser? How long will it take to completely vaporize 1 kg of water collected from the tap if an equally powerful heater is used for this purpose? Solution: 600 Mass of water heated every second = = 10 gm. 60 ∴The heat required to raise the temperature of 10 gm of water from 10 0 C to 90 0 C in each second = 10×1× (90 −10) = 800 cal = 800 × 4.2 J = 3360 J [Specific heat of water = 1 cal/(gm. 0 C )] ∴The electric power = 3360 J/s = 3360 W = 3.36 kW. [1 W = 1 J/s]

The temperature of tap water = 90 0 C . The heat required for the complete vaporization of 1 kg of water at 90 0 C consists of two parts: (i) The heat needed to raise the temperature of water to boiling point (100 0 C ) = 1000×1× (100 − 90) = 10000 cal = 10 kcal (ii) The latent heat for boiling 1 kg of water = 1000 × 540 = 540000 cal = 540 kcal

Total required heat for complete vaporization = (10 + 540) = 550 kcal = 550× 4.2 J = 2310 J. 2310 Thus the time taken for this = = 0.6875 sec. 3360 Example 7: A container contains some water. If now some vapour is being taken out from this container with the help of a pump, some of the water begins to turn ice due to rapid vaporization. How much water could be converted into ice by this method? The latent heat of freezing of water = 80 cal/gm and the latent heat of vaporization = 540 cal/gm. Solution: Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 15

Suppose, the initial mass of water = m gm and out of this water m1 gm turns ice and m2 gm is evaporated.

∴ m = m1 + m2

The latent heat removed from m1 gm water for converting it to ice = m1 ×80 = 80.m1 cal.

The heat required for m2 gm water for vaporization = m2 × 540 = 540.m2 cal. Here, we can write 4 540.m = 80.m Or, m = m 2 1 2 27 1 4 31 27 ∴ m = m + m = m Or, m = m 1 27 1 27 1 1 31 27 Thus a fraction of initial mass of water may be converted to ice. 31 0 Example 8: A piece of copper, which weighs100 gm, is heated up to 100 C and then dropped into a 100 gm copper calorimeter. The calorimeter contains 40 gm mixture of some water and some ice. Find out the initial mass of ice in the mixture if the final temperature becomes 10 0 C . The specific heat of copper = 0.09 and the latent heat of melting of ice = 80 cal/gm. Solution: Let us suppose, the calorimeter contains x gm of ice. ∴There is (40 − x ) gm water in the calorimeter at 0 0 C . The heat rejected by the copper piece = 100× 0.09× (100 −10 ) = 810 cal. The total absorbed heat has the following parts: (i) The heat absorbed by ice at 0 0 C for melting = x ×80 = 80.x cal (ii) The heat absorbed by the water from melted ice at 0 0 C to get heated up to 10 0 C = x ×1× (10 − 0 ) = 10.x cal. (iii) The heat absorbed by (40 − x ) gm water at 0 0 C to get heated up to 10 0 C = (40 − x) ×1× (10 − 0) = (400 −10.x ) cal. (iv) The heat absorbed by the calorimeter = 100× 0.09× (10 − 0 ) = 90 cal. ∴The total absorbed heat in the calorimeter = 80.x +10.x + (400 −10.x) + 90 = 80.x + 490 According to calorimetric principle, the total heat rejected = the total heat absorbed. ∴ 80.x + 490 = 810 Or, 80.x = 320 Or, x = 4 ∴The mixture in the calorimeter contains 4 gm ice.

0 Example 9: A calorimeter contains a mixture of 250 gm water and 200 gm ice at 0 C . The water equivalent of the calorimeter is 50 gm. Now 200 gm steam at 100 0 C is passed through this mixture. (i) What will be final temperature of the mixture? (ii) What will be the final weight of the mixture if the final temperature is 100 0 C ? The latent heat of melting of ice = 80 cal/gm and the latent of steam = 540 cal/gm. Solution: The steam at 100 0 C is condensed into water at 100 0 C . Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 16

(i) In this case, we assume that all the steam is converted to water. Let the final temperature of the mixture becomes t 0C and this temperature should be something between 0 0 C and 100 0 C . The total rejected heat consists of two parts: The heat rejected by steam at 100 0 C when it becomes water at 100 0 C = 200× 540 = 10800 cal. The further heat rejected by the water at 100 0 C as it cools down to t 0C = 200×1× (100 − t) = (20000 − 200.t ) cal. ∴Total rejected heat = 10800 + (20000 − 200.t ) = 30800 − 200.t cal.

The total heat absorbed in the system, consists of the following parts: The heat absorbed by ice at 0 0 C for melting = 200×80 = 16000 cal. The heat absorbed by the water from melted ice at 0 0 C to become t 0C = 200×1× (t − 0) = 200.t cal. The heat absorbed by 250 gm water initially at 0 0 C = 250 × t = 250.t cal. The heat absorbed by the calorimeter = 50× t = 50.t cal. ∴Total absorbed heat = 16000 + 200.t + 250.t + 50.t = 16000 + 500.t cal. According to calorimetric principle, 148 16000 + 500.t = 30800 − 200.t Or, 700.t = 14800 Or, t = = 21.14 7 ∴The final temperature of the system is 21.14 0 C . In the above case it has been assumed that all the steam is converted into water. (ii) In this case, we assume some steam escapes into surroundings. Let x gm steam could be converted into water. As the final temperature of the mixture is assumed to be 100 0 C , the total rejected heat will be by x gram steam at 100 0 C when it becomes water at 100 0 C = x × 540 = 540.x cal. The total heat absorbed in the system consists of the following: The heat absorbed by ice for melting = 200×80 = 16000 cal. The heat absorbed by the water from melted ice at 0 0 C to become 100 0 C = 200×100 cal. The heat absorbed by the water initially at 0 0 C (mixed with ice) = 250×100 cal. The heat absorbed by the calorimeter of water equivalent 50 gm = 50 ×100 cal. ∴ The total heat absorbed in the system = 16000 + (200 + 250 + 50) × 100 = 6666000 cal. Thus we can write, 66000 540.x = 66000 Or, x = = 122.2 540 ∴The final mass of the mixture = mass of water + mass of water from melted ice + mass of water from condensed steam = 250 + 200 +122.2 = 572.2 gm.

Note: If the mass of the steam to be condensed is different than the amount (572.2 gm), the final temperature of the mixture will be different from 100 0 C . Notice that in the first part of Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 17 the problem, we assumed that entire steam was converted (no loss) into water and thus the final temperature was something smaller than 100 0 C .

0 Example 10: A 50 gm metal piece at 10 C is left under the flow of steam and 0.8 gm of steam is condensed. What is the specific heat of the metal? Given, the latent heat of steam for condensation = 540 cal/gm. Solution: Let the specific heat of the metal = S . The temperature of steam = 100 0 C . Hence, the temperature of the metal piece will eventually be 100 0 C . The heat absorbed by the metal piece = 50× S × (100 −10 ) = 4500.S cal. The heat rejected by steam for condensation = 0.8× 540 cal. Q The heat absorbed = the heat rejected 0.8× 540 ∴ 4500.S = 0.8× 540 Or, S = = 0.096 4500 ∴The specific heat of the metal = 0.096.

0 0 Example 11: In a boiler, 10 kg water has to be heated from 20 C to 80 C in one hour with the help of superheated steam at 150 0 C passing through a conducting pipe placed inside the boiler. Steam condenses into water at 90 0 C inside the pipe and this water flows out of the pipe. How much steam is required for this purpose in each hour? The specific heat of steam = 1 cal/(gm. 0 C ) and latent heat of steam = 540 cal/gm. Solution: The heat required in each hour = 10×103 ×1× (80 − 20 ) = 60×104 cal. This heat is received from the passing steam. Let x kg steam is required each hour. ∴The total heat rejected by steam = the heat rejected by the steam at 150 0 C to become steam at 100 0 C + the latent heat removed by steam at 100 0 C to condense water at 100 0 C + heat rejected by the water at 100 0 C to cool down at 90 0 C = x ×103 ×1× (150 −100) + x ×103 × 540 + x ×103 ×1× (100 − 90) = x ×103 × (50 + 540 +10) = x ×103 × 600 cal. ∴We can write, x ×103 × 600 = 60×104 Or, x = 1

∴Steam required at each hour = 1 kg.

Example 12: Water vaporizes from an earthen pot at the rate of 1 gm/min. If the water equivalent of the pot is 500 gm and the pot contains 9.5 kg water, how long does it take for the water pot to cool down from 30 0 C to 28 0 C ? The latent heat of vaporization of water = 580 cal/gm. Solution: Let us suppose, ∆m gm water gets vaporized during the required time. ∴The latent heat taken by the evaporated water for vaporization = ∆m × 580 cal. This absorbed heat is taken from the water and the earthen pot. Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 18

The heat rejected by water and the earthen pot = (500 + 9500) ×1× (30 − 28 ) = 20000 cal. 20000 ∴ ∆m × 580 = 20000 Or, ∆m = = 34.48 580 Now, 34.48 gm water is evaporated at the rate of 1gm/min. 34.48 ∴The time taken for the water pot to cool down = = 34.48 min. 1

Alternative Method:

This problem can be attempted by the method of Calculus. Let the amount of water in the earthen pot at any time is m and the temperature is θ . If now dm amount of water vaporizes and the temperature goes down by dθ , we have the heat taken by the vaporized water = the heat rejected by the water pot and the water: L.dm = (W + m).S.dθ , where L = the latent heat of vaporization; S = specific heat of water; W = water equivalent of the earthen pot. L dm From the above, = dθ S (W + m) m1 d(W + m) S θ1 Integrating on both sides, = dθ ∫ (W + m) L ∫ m0 θ0

[ m0 is the initial mass of water when temp. = θ 0 ; m1 is the mass when temp. = θ1 ]

m1 S  W + m1  S Or, ln(W + m) = (θ1 −θ 0 ) Or, ln  = (θ1 −θ 0 ) m0   L W + m0  L S   ()θ1 −θ0 W + m1 L x Or,   = e = e W + m0  S 1× (28 − 30) Here, the term in the exponential, x = (θ −θ ) = = − 0.003 L 1 0 580 0 0 0 [We put S = 1 cal/(gm. C ), L = 580 cal/gm, θ1 = 28 C and θ 0 = 30 C . ∴ x is dimensionless]

 W + m1  −0.003448 x ∴   = e ≅ 1− 0.003448 [If x <<1, we can write e ≈ 1+ x ] W + m0 

Thus we have, (W + m1 ) = (1− 0.003448) × (W + m0 )

∴ ∆m = (m0 − m1 ) = 0.003448× (W + m0 )

∴The amount of water vaporized, ∆m = m0 − m1 = 0.003448× (500 + 9500 ) = 34.48 gm.

[Note: The two methods will give same result as long as the exponent x in the exponential is small enough (Note the end expression of ∆m .). The exponent x small S when the temperature difference (θ −θ ) happens to be small; the other factor is 1 0 L usually much smaller than 1.] Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 19

0 Example 13: A solid substance of 10 gm is taken whose temperature is −10 C . The heat required to raise its temperature to − 20 C is 64 cal. The substance remains solid at this temperature. The amounts of heat required to raise the temperatures of the substance in liquid state up to 1 0 C and 3 0 C are 880 cal and 900 cal, respectively. Suppose the specific heats of the substance in the solid and liquid states are S1 and S 2 ( S1 ≠ S 2 ), respectively. Find their values. Also show that the latent heat of melting L and the melting point tm are related by L = 79 + 0.2.tm . Solution: The heat required to raise the temperature of the substance from −100 C to − 20 C in solid state = 10× S1 ×[−2 − (−10)] = 80.S1

∴ 80.S1 = 64 Or, S1 = 0.8 The heat required to raise the temperature of the substance from 1 0 C to 3 0 C in liquid state = 10× S 2 × (3 −1) = 20.S 2

∴ 20.S 2 = 900 − 880

Or, 20.S 2 = 20 Or, S 2 = 1

If the melting point of the solid substance is tm , the heat required to raise the temperature of the substance from −100 C to the melting point; then allow it to melt and finally raise the temperature of the substance in liquid state up to 1 0 C =

10× S1 ×[tm − (−10)] +10.L +10× S2 × (1− tm ) = 10.[L +10.S1 + S 2 − tm (S2 − S1 )] ∴ According to question,

10.[L +10.S1 + S 2 − tm (S2 − S1 )] = 880

Or, L +10.S1 + S2 − tm (S 2 − S1 ) = 88

Or, L +10× 0.8 +1− tm (1− 0.8) = 88 Or, L + 8 +1− tm × 0.2 = 88

∴ L = 79 + 0.2.tm

Example 14: Heat is supplied at the rate of 200 cal/min to a 100 gm solid body. It is seen that (i) the temperature of the body increases from − 50 C to 00 C in 1.25 min; (ii) the temperature remains at 00 C for 40 min; (iii) next, the temperature increases at the rate of 2 0 C /min and reaches up to 100 0 C ; (iv) the temperature remains fixed at 100 0 C for 135 min and at this time the quantity of the material reduces to half. From the above set of observations can you determine the heat constants of the material of the body? Ignore any kind of heat losses. Solution: (i) The total heat absorbed by the body for the increase of temperature from − 50 C to 0 0 C = 200×1.25 = 250 cal. If the specific heat of the solid material = S1 , we can write

100× S1 ×[0 − (−5)] = 250 Or, 500.S1 = 250 ∴ S1 = 0.5

Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 20

(ii) As the temperature remains constant although the heat is supplied, we can conclude that there is a phase change occurring at this stage. If the latent heat of melting is L1 cal/gm, 40× 200 100.L = 40× 200 Or, L = = 80 1 1 100 ∴ The latent heat of melting of the material = 80 cal/gm.

(iii) The material is in liquid phase at this stage. If the specific heat in the liquid phase is 0 S 2 , the heat required to raise the temperature at 100 C = 100× S 2 × (100 − 0) =

10000.S 2 cal. 100 Time taken to raise this temperature = = 50 min. 2 ∴Total heat supplied at this stage = 200 × 50 = 10000 cal.

Thus we can write, 10000.S 2 = 10000 ∴ S 2 = 1

(iv) As the temperature remains fixed for 135 min, this indicates that there is again a phase transition occurring. The material reduces by half which means half of the liquid, that is 50 gm is vaporized. Total heat absorbed by the system at this stage = 135× 200 = 27000 cal.

If the latent heat of vaporization is L2 , we can write, 27000 50× L = 27000 Or, L = = 540 2 2 50 ∴The latent heat of vaporization = 540 cal/gm.

0 Example 15: A 7.5 gm copper piece at 27 C is dropped into boiling liquid oxygen (boiling point -183 0 C ). For this, the vaporized oxygen occupies a volume of 1.83 litre at 20 0 C and at a pressure of 750 mm. What is the latent heat of vaporization of oxygen? The specific heat of copper = 0.08; the density of oxygen at N.T.P. = 1.429 gm/litre. Solution: The volume, pressure and temperature of the vaporized oxygen are

V1 = 1.83 litre, P1 = 750 mm, T1 = 273 + 20 = 293 K.

The pressure and temperature at N.T.P. are P2 = 760 mm, T2 = 273 K. Let the volume at

N.T.P be V2 litre.

P1V1 P2V2 P1V1T2 750×1.83× 273 We can write, = Or, V2 = = = 1.68 litre T1 T2 P2T1 760× 293 ∴The mass of the vaporized oxygen = 1.68×1.429 gm. If the latent heat of vaporization of oxygen is L cal/gm, the heat absorbed by oxygen for vaporization = 1.68×1.429 × L cal. The heat rejected by the piece of copper = 7.5× 0.08×[27 − (−183 )] = 7.5× 0.08× (27 +183) = 7.5× 0.08× 210 cal. ∴1.68×1.429× L = 7.5× 0.08× 210 Or, L = 52.5 ∴The latent heat of vaporization of oxygen = 52.5 cal/gm. Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 21

Example 16: What will happen if 64800 cal of heat is removed from 100 gm steam at 100 0 C . The latent heat of steam = 540 cal/gm; the latent heat for melting of ice = 80 cal/gm. Solution: As the latent heat of steam = 540 cal/gm. Hence to convert 100 gm water at 100 0 C into 100 gm steam at 100 0 C , we require 100 × 540 = 54000 cal of latent heat. ∴If we now remove 54000 cal of heat from the 100 gm steam, we get back 100 gm water at 100 0 C . Now if we want to reduce the temperature of 100 gm water from100 0 C to 0 0 C , we have to remove 100×1×100 = 10000 cal of heat. ∴Total heat removed at the two steps = 54000 + 10000 = 64000 cal. We have to further remove (64800 – 64000) = 800 cal of heat. If we remove this heat, the water at 0 0 C starts freezing. The latent heat of freezing of ice = 80 cal/gm. 800 ∴We can freeze = 10 gm of water, which means we shall have 10 gm ice at 0 0 C 80 and (100 – 10) = 90 gm water at 0 0 C .

0 Example 17: Divide 1 kg water at 5 C in such a way that the heat rejected for freezing of one part into ice at 0 0 C is equal to the amount that can be used to vaporize another part into steam. The latent heat of ice and steam are 80 cal/gm and 540 cal/gm, respectively. Solution: Let the mass of one part of water = x gm. The mass of other part = (1000 − x ) gm. The quantity of rejected heat for freezing x gm water at 5 0 C into ice at 0 0 C = x ×1× (5 − 0) + x ×80 = 85.x cal. The heat required to convert (1000 − x) gm water at 5 0 C to steam = (1000 − x)×1× (100 − 5) + (1000 − x) × 540 = (1000 − x) × 635 cal. ∴We can write, 85.x = (1000 − x) × 635 Or, 85.x + 635.x = 635000 635000 Or, x = = 881.9 gm 720 ∴The amount of 1 kg water has to be divided into two parts: 881.9 gm to be converted to ice and the (1000 – 881.9) = 118.1 gm to be converted to steam.

0 0 Example 18: An electric kettle takes 15 min for heating some water from 0 C to 100 C and it takes 80 min to vaporize the water completely. What is the latent heat of vaporization? Solution: Let the kettle supplies heat at the rate of H cal/min and the amount of water in the kettle is m gram. ∴ We can write, Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 22

H 100 20 for heating the water, 15× H = m.(100 − 0 ) Or, = = and m 15 3 for vaporization, 80.H = m × L [ L = latent heat of vaporization] H 20 Or, L = 80. = 80× = 533.3 m 3 ∴ The latent heat of vaporization is 533.3 cal/gm.

Discussions of a few Questions

Q.1 Some amount of water at 100 0 C and an equal amount of steam at 100 0 C - which of them is more dangerous in terms of burning? 0 Ans. One gram of water at 100 C requires 540 cal of heat (latent heat of boiling) in order to be converted to one gram of steam at 100 0 C . Thus steam at 100 0 C contains more heat than water at 100 0 C . Hence steam is more dangerous in burning our skin.

Q.2 Ice at 0 0 C is more effective than water at 0 0 C for cooling – why? Ans. An amount of 80 cal heat (latent heat) has to be removed from 1 gm of water at 0 0 C to be converted to 1 gm of ice at 0 0 C . Thus every 1 gm of ice at 0 0 C contains 80 cal of heat less than every 1 gm of water at 0 0 C . Therefore, ice at 0 0 C is more effective in cooling than water at 0 0 C .

Q.3 What do you understand by ‘melting point of ice is 0.0073 0 C ’? 0 Ans. The melting point of ice is 0 C at normal atmospheric pressure. Ice is such a substance whose melting point decreases when a high pressure is applied on it and conversely, the melting point increases when the pressure on it is reduced. We understand that the ice is kept at a pressure below the normal atmospheric pressure. In this state ice starts melting at 0.0073 0 C and the temperature remains fixed until the entire ice melts.

Q.4 A hole is made inside a chunk of melting ice. Now water is poured into the hole. Will that water be converted to ice? 0 Ans. The answer is no. The temperature of melting ice is 0 C . The water in contact with melting ice, initially at a temperature higher than 0 0 C , will reject heat and this heat melts some ice. If the chunk of ice is big enough, the temperature of water in the whole eventually turns to 0 0 C . But to convert this water at 0 0 C latent heat of 80 cal/gm is needed to be removed which the ice at 0 0 C is not able to absorb (The heat exchange is not possible as both ice and water being at 0 0 C .) . Thus the water remains at 0 0 C .

Q.5 If steam at normal atmospheric pressure is passed through water kept at a beaker, will it be possible to boil that water? Ans. The water will not boil. The temperature of steam at normal atmospheric pressure is 100 0 C . When this steam is passed through water, the steam will reject heat which the Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 23 water absorbs and gets heated. Eventually, the temperature of water turns to 100 0 C . But for the water to boil, it requires 540 cal/gm of heat (latent heat) more to be absorbed. This heat can not be absorbed by water from the passing steam as the water and steam are now both at 100 0 C . Thus the water can not boil and remains at 100 0 C .

Q.6 An astronomer on the moon pours water at 20 0 C into a beaker from a thermo flask Explain what happens to this water. Ans. There is no atmosphere on the moon. Thus the atmospheric pressure is zero on the moon. Thus there will be no atmospheric pressure on the surface of water in the beaker except the pressure exerted by the water vapour. In this situation, the water will start boiling (boiling point of water will come down) and the latent heat of vaporization will be taken from the water itself. Thus the temperature will be reduced which may go down to freezing point and it is possible for the water to become ice.

Q.7 Why is it advisable not to immerse the thermometer bulb inside the boiling water when determining the boiling point of water? Ans. In case of water not absolutely pure, the boiling point may increase due to some impurity dissolved in it. But the vapour coming out this boiling water is pure, no impurity is mixed with this. Thus the temperature of the vapour equals to the boiling point of pure water. Thus the thermometer should be placed inside the emerging vapour rather than in boiling liquid.

Q.8 Why the mercury column in a thermometer comes down if the mercury bulb is wrapped with a wet cloth? If ether is used instead of water will the rate of coming down of mercury column change? Ans. Water from the wet cloth evaporates and absorbs heat (latent heat of evaporation) from the mercury in the bulb. Thus the temperature of mercury comes down and thus the column of mercury contracts. If instead of water ether is used, the rate of evaporation will be faster as ether is a volatile substance. Thus the rate of cooling will be faster.

Q.9 Why does evaporation cause cooling? Ans. When a liquid evaporates into vapour, it requires latent heat for this change of phase. If the heat is not supplied from outside the liquid takes it from its own body and from the surroundings. This makes the liquid and the surroundings cool.

Q.10 Why does it take more time to reach the water from 0 0 C to the boiling point at 100 0 C than the time to complete the vaporization of water by boiling even if the heat is supplied at the constant rate? 0 Ans. The heat required for 1 gm water to raise its temperature by 1 C is 1 cal. Thus the heat required to raise the temperature of water from 0 0 C to 100 0 C is 100 cal. But to vaporize 1 gm water from its boiling point (100 0 C ) it takes 537 cal of heat. Hence the heat required is 5.37 times more in the later case. Therefore, as the heat is supplied at the Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 24 constant rate, it takes 5.37 times more time to vaporize 1 gm water from its boiling point than the time taken by 1 gm water to raise its temperature from 0 0 C to its boiling point.

Q.11 An aluminium and a lead ball of equal mass, kept at room temperature, are hanged from threads. Now they are immersed in melted naphthalene (in equilibrium with solid naphthalene in a container) for a while and then taken out. The balls have got coating of naphthalene on their surface. When the balls are weighed, it is seen that the mass of deposited naphthalene on the aluminium ball is more than that of the lead ball. Why is this difference? Explain. Ans. The two balls have equal mass, same initial and final temperatures. Thus the increase of temperature for the two balls is the same. But the heat absorbed by the balls from melted naphthalene would depend on the specific heat of the material of the balls as the absorbed heat = mass×specific heat×temperature increase. The balls absorb latent heat from the melted naphthalene when they are immersed in it. This makes the naphthalene to lose heat and gets solidified on the surface of the metal balls. Hence is the deposition. More the absorbed heat the more is the deposition. The absorbed heat is more in case of higher specific heat. The specific heat of aluminum must be more than the specific heat of lead as the deposition on the aluminium ball is more.

Q.12 The readings of a thermometer are different when some drops of ether are put on the mercury bulb of a thermometer than when the mercury bulb itself is immersed in a ether bottle – explain why. Ans. In the first case, ether evaporates quickly and takes the latent heat of evaporation from the mercury in the bulb. Thus the mercury column comes down. But in the second case, the mercury bulb is immersed in the ether. The ether, in touch with the bulb inside the liquid, can not evaporate (evaporation occurs on the surface only). Thus the mercury column in the bulb does not come down.

Q.13 Is it possible to boil water at room temperature? Ans. We know the boiling point comes down if the air pressure on the exposed surface of a liquid can be decreased. At high altitude, owing to the low atmospheric pressure, water boils at a much lower temperature than 100 0 C . If the water can be taken in a vacuum (air pressure tends to zero), it is possible to boil it at room temperature. Note: Get an injection syringe (without the needle). Suck a little amount of water into the syringe. Now cover the opening on the syringe somehow with finger or something else and then pull back on the plunger as hard and as fast as possible. This will create a partial vacuum inside the syringe, and it is possible to see the water in it boiling. (Reference: http://invsee.asu.edu/ed/phase)

Q.14 Ammonia gas can be converted to liquid by applying pressure at normal temperature; but oxygen gas can not be converted to liquid at the same temperature – why? [J.E.E.] Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 25

Ans. The normal boiling point of a liquid is the temperature at which its vapour pressure is equal to one atmosphere. When the external pressure is greater than one atmosphere, the boiling point of the liquid is greater than its normal boiling point. Therefore, it is possible to boil the liquid at a higher temperature than its boiling temperature; which means the liquid remains liquid up to a higher temperature. The boiling point of liquid ammonia is -33 0 C . It is possible to raise the boiling point above the normal room temperature of 27 0 C with the application of high enough pressure. But the boiling point of oxygen is -183 0 C , which is much below the normal room temperature. Thus it is difficult to raise the boiling temperature of oxygen above the room temperature and thus to get liquid oxygen at room temperature by applying external pressure.

Q.15 Is it possible to heat a solid beyond its melting point? Ans. It is difficult and almost impossible to heat a solid above its melting point because the heat that is absorbed by the solid at its melting point is used to convert the solid into a liquid. The temperature does not change until the entire solid melts. It is possible, however, to cool some liquids to temperatures below their freezing points without forming a solid. When this is done, the liquid is said to be a supercooled liquid.

Questionnaire

Very Short Questions: Mark: 1

(Answer in one or two words)

1. What is termed as the heat absorbed by a substance at constant temperature during a phase change? [Latent heat] 2. What is the unit of latent heat in S.I. system? [Joule/kg] 3. What is known to be the melting of ice by applying extra pressure and then refreezing of this to solid ice by withdrawing the pressure? [Regelation] 4. What is the process by which the vaporization of a liquid takes place from its surface at any temperature? [Evaporation] 5. What is the process by which the vaporization of a solid takes place from its surface at any temperature? [Sublimation] 6. Can the boiling point and the freezing point of a substance be equal? [yes, at triple point]

(Fill in the Blanks)

1. The latent heat of vaporization is approximately ------cal/gm. [80] 2. The latent of boiling of water is approximately ------cal/gm [540] 3. The melting point of ice ------if the pressure on it is increased. [increases] 4. The boiling point of a solution is ------than the boiling point of the solvent. [more] Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 26

5. The pressure cooker cooks faster as the ------increases due to extra pressure. [boiling point]

(Multiple Choice type)

1. The freezing point of water is 0 0 C . If now some salt is mixed with the water the freezing point will be (a) 0 0 C , (b) more than 0 0 C , (c) less than 0 0 C (d) the water will never freeze. [(c)] 2. Cooking on a hill resort or mountain top takes longer time. This is because (a) the temperature of atmosphere is less, (b) the humidity is less, (c) the atmospheric pressure is lower, (d) atmospheric pressure is higher. [(c)]

3. The relation between the melting point and freezing point of a pure substance is that (a) the two temperatures are equal, (b) melting point > freezing point, (c) melting point < freezing point, (d) melting point may be more or less than the freezing point. [(a)] 4. The melting point of ice in S.I. unit is (a) 0.336 J/kg, (b) 3.36×105 J/kg, (c) 80 kcal/kg, (d) 80000 J/kg. [(b)]

5. If 10 gm ice at 0 0 C is dropped into 80 gm water at 20 0 C the final temperature of the mixture will be (a) 0 0 C , (b) 2 0 C , (c) 10 0 C , (d) 18 0 C [(c)]

6. For an electrical fuse (a) alloy material is used as its melting point is low, (b) alloy material is used as its melting point is high, (c) alloy is never used as its melting point is low, (d) alloy is never used as its melting point is high. [(a)]

7. There is a strong wind blowing. Because of this the rate of evaporation (a) decreases, (b) increases, (c) may increase or decrease, (d) remains same. [(b)]

8. Pressure cookers allow food to cook more rapidly due to (a) lower melting-freezing points, (b) higher boiling points, (c) higher melting- freezing points, (d) lower boiling points [(b)]

Short Questions: Marks: 2

1. What do you mean by phase change? Explain with examples. 2. Define the melting point of solid. 3. What are the melting and freezing of a substance? 4. Define freezing point of a material. 5. Why do the water pipes burst in winter in cold countries? 6. Are the melting point and the freezing point of a material same? [H.S. ’98, ‘97] Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 27

7. What do you mean by the melting point of lead by 327 0 C ? 8. Discuss the effect of pressure on melting point. [H.S. ‘05] 9. Discuss the role of impurity on the boiling point of a liquid. 10. Why does food cook faster in a pressure cooker? 11. Is hard water good for cooking? 12. What is regelation? [H.S. ’94, ‘91] 13. How does the volume of a substance generally change during melting or during freezing? Discuss with examples if there are exceptions. 14. The volume increases when water freezes to become ice. What are the advantages and disadvantages of this? 15. What is latent heat of melting? [H.S. ’02, ’99, ‘92] 16. What are ‘sensible heat’ and ‘latent heat’? 17. What do you mean by the latent heat of melting of ice to be 80 cal/gm? 18. What is the latent heat of melting of ice in F.P.S. system and in S.I. system? 19. Why the latent heat of vaporization of a substance is more than the latent heat of melting? 20. What is evaporation? 21. What are the factors that influence the rate of evaporation? [H.S. ’96, ’94, ‘90] 22. Why does cooling result in due to evaporation? Explain with examples. 23. What do you mean by latent heat of evaporation? 24. What do you mean by the latent heat of steam to be 537 cal/gm? 25. What will be the change in the boiling point of a liquid when the pressure on its surface is reduced? 26. Show with an example that the boiling point of a liquid increases as the pressure on its surface is increased. [H.S. ‘93] 27. Is there any difference between boiling and evaporation? [H.S. ’01, ’99, ‘94] 28. Discuss in brief the working principle of a pressure cooker. [H.S. ’93, ‘90] 29. Discuss about a practical use of the effect of pressure on boiling point. [H.S. ‘95] 30. What is the boiling point of a liquid? What do you mean by the boiling point of benzene to be 80 0 C ? 31. How does the height of a place influence the boiling point of a liquid? 32. What is the effect of pressure on boiling point? [H.S. ’01, ’99] 33. What is the role of latent heat? 34. Which kinds of materials generally have the same freezing and melting points and which do not have them to be the same? 35. Why do the hot water pipes in cold countries develop crack more often than the cold water pipes? 36. Why do cracks appear on the mountain stones? [H.S. ’02, ’00] 37. Why do two pieces of ice join when they are pressed together? 38. Why is the electrical fuse made of an alloy rather than a pure metal? 39. Why does the melting point of ice come down if salt is mixed with it? 40. What is a freezing mixture? 41. Why is the ice at 0 0 C more effective than the water at 0 0 C to cool a substance? 42. Why is the steam at 100 0 C more capable of burning your skin than boiling water? Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 28

43. If water is poured in a hole of an ice chunk will the water turn ice? 44. A beaker contains water at room temperature. Will it be possible to convert the water into steam by passing steam through this? 45. What do you mean by the melting point of ice to be 0.0073 0 C in vacuum (no atmosphere). 46. In a place where the atmospheric pressure is zero, the water starts boiling at a lower temperature and the temperature of water comes down. Can you explain this? 47. Why the vessel is generally kept covered while cooking on an oven? 48. The water remains cool in an earthen pot than in a metal container in summer. Explain this scientifically. 49. Why do you feel comfort when you sweat and then sit under the fan? 50. If you take a few drops of ether on your palm, you feel a cooling sensation. Explain this scientifically. 51. When hot tea is poured over a plate from a cup or just spills from a cup over the ground, it cools faster. Explain this scientifically. 52. Cooking takes more time on a high altitude. Explain this. 53. Why does water boil at a temperature much lower than 100 0 C at Darjeeling (a hill resort)? [J.E.E. ‘97] 54. Water boils at 120 0 C in a pressure cooker. What is the reason behind this? [J.E.E. ‘00] 55. What are the things that influence the boiling point of liquid? 56. In cold countries, the lakes, rivers freeze in winter. But the sea water does not easily freeze. Explain this. 57. Is it possible for the water to boil at room temperature? Explain with reason. 58. In what conditions, it is possible to supply heat to a body but not changing its temperature. How is it possible to remove that heat? 59. Why the work done to vaporize a substance is more than the work done to melt that body from its solid state?

60. P Watt power has to be supplied to keep a substance of mass M kg in a molten state at its melting point. If the power source is removed, the substance freezes completely in t sec. Find the latent heat of melting of the substance. [I.I.T. ‘92]

61. Write four differences between boiling and evaporation. [H.S. ‘04]

62. How does the melting point of ice depend on the surrounding pressure? [H.S. ‘05] 63. What do you mean by boiling point? What is the effect of pressure on this? [H.S. (XI) ‘06]

Simple Problems: Marks: 2

1. How much heat is required to convert 100 gm water at 100 0 C into steam? Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 29

[H.S. ‘93] [Ans. 54000 cal.] 2. A quantity of 300 gm water at 25 0 C is added to 100 gm ice at 0 0 C . Find the final temperature of the system. [Ans. 0 0 C ]

3. A vessel contains 1 litre water. For lowering the temperature from 77 0 F to 50 0 F , how many blocks of ice are required if each block weighs 20 gm? [Ans. 8]

4. What will be the result if 69600 cal of heat is removed from 100 gm steam at 100 0 C ? The latent heat of steam = 540 cal/gm; the latent heat of ice = 80 cal/gm. [H.S. ’05; J.E.E. ‘03] [Ans. 70 gm ice and 30 gm water at 0 0 C ]

5. Hot air is blowing which is supplying heat at the rate of 5000 cal per minute. A 1000 kg ice chunk at 0 0 C is kept in an open box so that it also assumes 0 0 C . How long will it be possible to keep the temperature of the box at 0 0 C ? [Ans. 11.1 days]

6. A quantity of 10 gm ice at -8 0 C is to melt and then the temperature has to be raised to 50 0 C . How much heat is required? The specific heat of ice = 0.5; the latent heat of melting of ice = 80 cal/gm. [Ans. 1340 cal.]

7. How much heat is required to convert 2 gm ice at -2 0 C to steam at 100 0 C ? The specific heat of ice = 0.5; the latent heat of melting of ice = 80 cal/gm; the latent heat of steam = 540 cal/gm. [Ans. 1460 cal.]

8. How much heat is required to melt 200 gm tin at 32 0 C ? [The melting point of tin = 232 0 C ; the latent heat of melting = 14 cal/gm; the specific heat of tin = 0.055] [Ans. 5000 cal.] 9. If 5 gm ice at -10 0 C is mixed with 22 gm water at 30 0 C what will be the final temperature of the mixture? The specific heat of ice = 0.5. [Ans. 8.7 0 C ]

10. 100 gm water at 40 0 C is added with 100 gm ice at 0 0 C is taken. Find the final temperature of the mixture and also determine the proportions of ice and water. [Ans. 0 0 C ; 150 gm water, 50 gm ice] 11. A calorimeter of water equivalent 30 gm contains 270 gm water at 30 0 C . Now 10 gm ice at -10 0 C is dropped in it. What will be the final temperature of the mixture? The specific heat of ice = 0.5; the latent heat of melting of ice = 80 cal/gm. [Ans. 26.29 0 C ]

12. A calorimeter of water equivalent 5 gm contains 55 gm water at 20 0 C . If now 25 gm ice at 0 0 C is dropped in it, what will be the final temperature and what will be the proportion of the components in the mixture at this temperature? [Ans. 0 0 C ; 70 gm water, 10 gm ice] 13. A piece of 16 gm iron at 112.5 0 C is placed inside a hole in a chunk of ice and it is seen than 2.5 gm of ice melts. The latent heat of melting of ice = 80 cal/gm. What is the specific heat of iron? [Ans. 0.11] Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 30

14. A closed container contains 10 gm ice at 0 0 C . Now steam at 100 0 C is passed through this in such a way that the ice just melts. How much water will be there in the container in this situation? Given, the latent heat of melting of ice = 80 cal/gm, the latent heat of steam = 540 cal/gm. [Ans. 11.25 gm]

15. A container contains 5 gm ice and 100 gm water at 0 0 C . Now steam is passed through this mixture and the temperature is raised to 10 0 C . Find the amount of condensed vapour neglecting the water equivalent of the container and the loss of heat due to radiation. Given, the latent heat of condensation of vapour = 536 cal/gm and the latent heat of melting of ice = 80 cal/gm. [Ans. 2.316 gm]

16. Steam at 100 0 C is flown over a chunk of ice at 0 0 C . After some time, 225 gm of water is collected. The mass of the ice chunk is reduced from 850 gm to 650 gm. Find the latent heat of condensation of steam. [Ans. 540 cal]

17. A 100 gm copper calorimeter contains 50 gm ice and 300 gm water at 0 0 C . How much steam at 100 0 C has to be passed through this so that the temperature of the mixture and that of the calorimeter rises to 10 0 C ? The specific heat of copper = 0.095; the latent heat of condensation of steam = 537 cal/gm; the latent heat of melting of ice = 80 cal/gm. [Ans. 12.1 gm]

18. The surface of earth receives solar energy at the rate of 1400 Watt/m 2 . The solar energy that impinges on a lens of area 0.2 m 2 is completely focused on to a 280 gm ice cube. How much time will it take to melt the ice completely? The latent heat for melting of ice = 3.3×105 J/kg. [I.I.T. ‘97] [Ans. 5.5 min]

Harder Problems

1. The temperature of some steam is 110 0 C and the temperature of ice is -15 0 C . In a mixture of the two, it is seen that the system is in equilibrium as water at a temperature of 40 0 C . Find the ratio of initial masses of ice and steam. The latent heat of condensation of steam and that of melting of ice are 540 cal/gm and 80 cal/gm, respectively; the specific heat of ice and that of steam are 0.53 and 0.48, respectively. [Ans. ice : steam = 4.73:1] 2. At the time of rapid vaporization of 10 gm water at 20 0 C , some water gets lost as steam and the rest becomes ice. Determine the quantity of ice. Given, the latent heat for melting of ice = 80 cal/gm and the latent heat of vaporization = 536 cal/gm. [Ans. 8.7 gm] 3. A layer of ice at 0 0 C is formed over the surface of a water body. Now 6 cm high rain water is collected over this. The temperature of the water is 12 0 C and the density of ice is 0.917 gm/cc. What is the height of ice layer that melts? [Ans. 0.98 cm]

Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 31

4. The boiling point of a liquid is 80 0 C . The vapour from the boiling liquid is passed through 90 gm water kept in a calorimeter. The water equivalent of the calorimeter is 10 gm and the temperature of the water is 30 0 C . If the temperature of water rises to 35 0 C after 5 gm vapour is passed through, find the latent heat of vaporization of the liquid. The specific heat of the liquid is 0.4. [Ans. 82 cal/gm]

5. Heat is supplied at a constant rate to ice for 7 minutes. In the first 1 minute, the temperature of ice increases at a constant rate. For the next 4 minutes, the temperature remains unchanged. In the last 2 minutes, the temperature again rises at a constant rate. Explain these observations and determine the final temperature. The latent heat of melting of ice = 80 cal/gm. [Ans. 40 0 C ]

6. A few pieces of ice at -10 0 C are being heated in a beaker at a constant rate. The temperature increases to 0 0 C after 1 minute. For the next 16 minutes, the temperature remains constant at 0 0 C . The temperature starts increasing again and reaches to 100 0 C after 20 minutes. Find the specific heat of ice and the latent heat of melting of ice. [Ans. 0.5; 80 cal/gm] 7. A 100 gm aluminium container contains 200 gm ice at -20 0 C . Heat is supplied to this system at a rate of 100 cal/sec. What will be the temperature of the system after 4 minutes? The specific heat of aluminium = 0.2, the specific heat of ice = 0.5, the latent heat of melting of ice = 80 cal/gm. [Ans. 25.5 0 C ]

8. A 100 gm copper piece is heated to 100 0 C and then plunged into a copper calorimeter. The calorimeter contains 40 gm mixture of ice and water. If the final temperature is 10 0 C find the quantity of ice in the mixture. The specific heat of copper = 0.09; the latent heat of melting of ice = 80 cal/gm. [Ans. 4 gm]

9. A metal container is partly filled with 5 gm water and sealed and the system is immersed into liquid oxygen at -180 0 C for some time. Then the sealed container is quickly transferred into a copper calorimeter. The temperature of the water in the calorimeter comes down from 20 0 C to 10 0 C . The heat capacity of the calorimeter and the water inside is 100 cal/ 0 C . Find the mean specific heat of ice in between -180 0 C and 0 0 C . The water equivalent of the metal container is 1 gm and the latent heat of melting of ice = 80 cal/gm. [Ans. 0.4]

10. A copper block at 40 0 C is placed on a big ice block at 0 0 C . If it assumed that the heat from the copper block is absorbed by the ice block, what fraction of the height of the copper block will be immersed into the ice block in equilibrium? The density of copper and ice are 8.6 gm/cc and 0.32 gm/cc, respectively; the specific heat of copper = 0.092 and the latent heat of melting of ice = 80 cal/gm. [Ans. 0.43 fraction]

11. A calorimeter, whose water equivalent is 100 gm, contains 900 gm water at 30 0 C . If 400 gm ice at 0 0 C is plunged into this, how much ice will melt? Now in this system - (i) How much steam has to be sent so that all the ice melts? (ii) How much steam is required Fundamental Physics I by Dr. Abhijit kar Gupta (email: [email protected]) 32 to raise the final temperature to 30 0 C ? Latent heat of melting of ice = 80 cal/gm; latent heat of condensation of steam = 540 cal/gm. [Ans. 375 gm, (i) 3.125 gm, (ii) 72.13 gm] 12. A copper calorimeter weighs 100 gm and this contains 240 gm water at 20 0 C . Now the temperature turns to 10 0 C as some pieces of ice are dropped into this. Now the calorimeter weighs 370 gm. How much water was sticking with the ice? The latent heat of melting of ice = 80 cal/gm; the specific heat of copper = 0.1. [Ans. 2.5 gm]

13. Water and oil mixture is taken at 20 0 C . Now 8 gm ice at 0 0 C is dropped into this and the ice melts. The final temperature becomes 0 0 C . If the specific heat of oil is 0.5 and the total weight of the entire liquid at the end is 60 gm, find the proportion of oil and water in the initial mixture. The latent heat of melting of ice = 80 cal/gm. [Ans. oil = 40 gm; water = 12 gm] 14. Steam mixed with water at 100 0 C is passed inside a 10 kg empty container. The temperature of the container rises from 15 0 C to 60 0 C due to this and also 150 gm water is seen to be collected. What was the percentage of water in the steam-water mixture? The latent heat for condensation of steam = 540 cal/gm; the specific heat of the material of the container = 0.12. [Ans. 40.74%]

15. A 10 gm metal piece at 100 0 C is dropped into an ice-water mixture. It is seen that the temperature of the mixture does not change but the volume reduces by 0.1 cc. If the specific heat of ice is 0.92 find the specific heat of the metal. Assume that the density of water at 0 0 C = 1 gm/cc. [Ans. 0.092]

16. If the volume of 1 gm of ice at 0 0 C reduces by 0.091 cc when melts, how much of a metal of specific heat 0.1 and of temperature 60 0 C is required so that when it is dropped into a ice-water mixture in a calorimeter, the volume of the mixture reduces by 0.273 cc? [Ans. 40 gm]

======