Physics 17 Part I

Temperature and Heat

The terms on the right all mean the Heat same thing. The heat energy of a Heat energy substance is the sum of the kinetic Thermal energy energies of all of the atoms. Internal energy

Kelvin Temperature Scale The lowest-possible Kelvin temperature of a substance is T = 0 oK. There are no negative Symbol: T Kelvin temperatures. Units: “degrees Kelvin” (oK) ……………………………………………… 0 oK is called “absolute zero.” The average kinetic energy per atom is proportional to the Kelvin temperature. This temperature scale is also called “the Absolute Temperature Scale.” Example: If the average kinetic energy per atom is 2 x 10-20 J at 300 K, then it would be At “absolute zero,” the motion of all atoms twice that at twice the Kelvin temperature: ceases. Nothing has a temperature of absolute at 600 K, the average kinetic energy per zero. atom would be 4 x 10-20 J.

Water Freezes: 273 oK Water Boils: 373 oK

o Celsius Temperature Scale Absolute Zero: TC = -273 C o Water Freezes: TC = 0 C o Symbol: TC Water Boils: TC = 100 C Units: “degrees Celsius” (oC) Because there are 100 degrees separating the TC = T - 273 freezing and boiling points of water, this scale is often called the “Centigrade Temperature Scale.”

Absolute Zero: TF = 9/5 (-273) +32 Fahrenheit Temperature Scale” = -459.4 oF

o Symbol: TF Water Freezes: TF = 32 F o o Units: “degrees Fahrenheit” ( F) Water Boils: TF = 212 F

TF = 9/5 TC + 32

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Example A: A Fahrenheit Kelvin Temperature Changes thermometer and a Celsius thermometer are placed in a Symbol: ΔT substance. The Fahrenheit Units: “Kelvin degrees” (Ko) reading is 20 degrees higher Note: superscript comes after the “K” than the Celsius reading. What o o are the two temperatures? Example B: To = 273 K (0 C) T = 373 oK (100 oC) TF = TC + 20 ΔT = T - To o (9/5) TC + 32 = TC + 20 = 100 K o Solve for TC: = 100 C (Celsius degrees)

o TC = -15 C Kelvin temperature change = Celsius temperature change

Now get TF: Note: Kelvin degrees and Celsius degrees are not the same TF = (9/5)(-15) + 32 as degrees Kelvin and degrees Celsius. The former are = 5 oF units of temperature change, while the latter are temperatures.

40 Co is NOT a temperature; it is not the same as the temperature 40 oC

Non-Standard Alternative Energy Unit Solution:

The “physicist’s calorie” (lower-case “c”) 140 x 1000 x 4.19 = 5.87 x 105 J 1.0 calorie (cal) = 4.19 J (“small” calorie) How much work is this?

The “dieter’s Calorie” (upper-case “C”) W = Fx 1.0 Calorie (Cal) = 1000 cal (“large Calorie”) If a 100-kg person (220 lbs), who weighs 980 N, lifts himself to the top of a 598-meter Example B: How much exercise would be building , he would do work in the amount of necessary to “burn off” one chocolate-frosting W = 980 x 598 = 5.87 x 105 J. cupcake (140 Cal).

In what follows in our study of quantities of heat entering and leaving substances, the symbol Q (for “quantity”) will be used. If heat enters a substance, Q will be positive; if heat leaves, Q will be negative.

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Phase Changes

Three of the more common forms (states) of matter, called “phases” of matter, are solids, liquids, and gases. When a substance changes from one phase to another, a “phase change” has occurred.

When a phase change is occurring, the temperature of the substance does not change.

Freezing Water Melting Ice

Ice will not begin to melt until its temperature has been raised to its melting point, which is 0 oC. After that temperature is reached, additional heat will cause the ice to melt. For each gram of ice at 0 o C to be melted, 80 cal of heat must be removed.

The only substance we will study in phase This number, 80 cal/g, is called the “latent changes is H2O, whose various phases include water, ice, and water vapor (steam). heat of melting”

Water will not begin to become ice until its L = 80 cal/g temperature is lowered to the freezing point of water, which is 0 oC. After that temperature is This number is called the “latent heat of reached, further removal of heat from the melting,” and is the same as the latent heat of water will causes ice crystals to form. To freezing. create one gram of ice from one gram of o water at 0 oC, 80 cal of heat must be removed. Note: as the ice at 0 C is melting, the temperature of the melting ice and water This number--80 cal/gram--is called the mixture does not change: it remains at a o “latent heat of freezing.” temperature of 0 C.

L = 80 cal/g Temperature doesn’t change in a phase change. Note: as the water at 0 oC is freezing, the temperature of the freezing ice and water Example: What quantity Q of heat must be o mixture does not change: it remains at a removed from 200 grams of water at 0 C to o temperature of 0 oC. convert it to 200 grams of ice at 0 C?

Temperature doesn’t change in a phase Answer: Q = -mL change. = -200 g (80 cal/g) = -16,000 cal = -16 kilocalories (kcal) Negative signs associated with Q’s indicate that heat is removed.

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Vaporization-Condensation Example A: How many calories of heat must be added to 200 grams of water at 100 oC to convert it to 200 grams of steam at 100 oC?

Q = +mL = 200 (540) = 108,000 cal = 108 kcal

(positive because heat is added) Once water’s temperature has been raised to its boiling point of 100 oC, it will begin to o evaporate. Each gram of water at 100 C Example B: How many calories of heat must requires 540 cal of heat to undergo a phase be removed from 50 g of steam at 100 oC to change to water vapor (steam). convert it to 50 g of water at 100 oC?

This number is called the “latent heat of Q = -mL vaporization.” = -50 (540) = -27,000 L = 540 cal/g = -27 kcal

The reverse process is condensation. To (negative because heat is removed) condense one gram of water vapor (steam) at 100 oC, 540 cal must be removed.

L = 540 cal/g

This number is called the “latent heat of condensation.”

Specific Heat Capacity If heat is added or removed from a substance whose phase is not changing, then a temperature change will occur. In what follows, temperature changes will be symbolized as ΔT, and will be measured in Celsius degrees (Co).

All substances have a property called “specific heat capacity,” symbolized as c, and Substance c o measured in units of cal/g-C . The quantity Q cal/g-Co of heat that must be removed or added to m Water 1.0 grams of a substance in order to cause a Ice 0.5 temperature change ΔT is Steam 0.5 concrete 0.2 Q = mcΔT aluminum 0.2

bone 0.1

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Re-write the equation above as: Example A: 900 cal of heat is added to 100 grams of water, and the same amount of heat ΔT = Q/mc is added to the same mass of concrete. Calculate and compare the temperature Because c is in the denominator, the larger the changes of each substance. specific heat capacity c of a substance, the smaller will be the temperature change when Water: ΔT = 900 / (100 x 1.0) heat is added or removed, compared to other = 9 Co substances with smaller heat capacities, and Concrete: ΔT = 900 / (100 x 0.2) vice-versa. = 45 Co The specific heat capacity of concrete is only one-fifth that of water, so concrete’s temperature rise is five times greater.

The greater the capacity of a substance to absorb heat, the less effect heat has on it. Water has a comparatively high specific heat capacity; almost everything else heats up or cools down more dramatically than water does.

Example B: How much heat must be added Example C: How much heat must be to 100 grams of water at 30 oC to raise its removed from 60 grams of ice at -30 oC to temperature to 70 oC? cool it down to -90 oC?

ΔT = 70 - 30 ΔT = -90 - (-30) = 40 Co = -60 Co Q = mcΔT Q = mcΔT = (100 g) (1.0 cal/g-Co) 40 Co = 60 (0.5) (-60) = 4000 cal = -1800 cal

Note that the sign of Q appears automatically; if ΔT had been negative, then Q would have been negative.

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Example A: See figure below. Q1 = 100 (0.5) (30) = 1500 Q2 = 100 (80) = 8000 o 100 grams of ice at -30 C is to be converted Q3 = 100 (1.0) (40) = 4000 to 100 grams of water at 40 oC. How much heat must be added to the ice to accomplish Q = Q1 + Q2 + Q3 this? = 13,500 cal = 13.5 kcal

Mixture Problems When two substances at different Example B: Sixty grams of aluminum at temperatures are mixed together, one of them 75 oC is added to 300 g of water at 30 oC. gains heat, while the other loses heat. Whatever one gains, the other loses: What is the equilibrium temperature?

Q1 + Q2 = 0 Q1 = 60 (0.20) (T -75) Q2 = 300 (1.00) (T -30) When “equilibrium” is reached, both substances will be at the same temperature. Q1 + Q2 = 0 Get T = 31.7 oC

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