Method of Mixtures

If a hot substance is added to, or, combined with, another substance which is at a lower temperature (ie it is “less hot” than the original substance, or perhaps colder) we would expect that the final temperature would be somewhere in between the two original temperatures.

Consider; If a mass of hot water is added to an equal mass of cold water we expect to end up with warm water or water that is tepid.

Note: the final temperature of the mixture is affected by; ▫ the mass of each substance involved in the combination ▫ the specific heat of each substance involved in the combination ▫ the initial temperature of each of the substances

This is consistent with the law of Conservation of Energy which can be simplified to; Heat Loss = Heat Gain In effect this states that “the amount of heat lost is equal, but opposite to, the amount of heat that is gained”. Using the symbols we have seen before we can express this as;

Q (substance A) Loss = Q (substance B) Gain Now we can use the formula Q = mcΔT and substitute it into the preceding equation. However, first we must be sure of the use which the symbol ΔT signifies. At the simplest level it represents the difference in temperature and is calculated by; o Tf – Ti where; Tf represents the final Temperature (in C or K) Ti represents the initial Temperature (in oC or K) If we now consider heat gain by a substance we expect the temperature of that substance to increase. In contrast, heat loss implies that the temperature of that substance would decrease.

What effect would this have on the equation Q = mcΔT? Note that Heat Loss is “the opposite” of Heat Gain and can be represented by the

“opposite” sign ie Q = - mc(Tf – Ti) which can also be expressed as mc(Ti – Tf)

Heat Gain: ΔT = Tf - Ti Heat Loss: ΔT = Ti - Tf

Now if we put all of this information together we get the following relationship Heat Loss = Heat Gain

mAcA(TiA – Tf(A+B)) = mBcB(Tf(A+B) – TiB)

Lazarov 2006 Consider the following examples.

1. A 0.5 kg block of aluminium at 100oC is placed in 1.0 kg of water at 20oC. What will be the final temperature of the aluminium and water at equilibrium assuming that there is no external heat gain or loss and no water is converted into steam?

Al Heat gain by H2O = Heat loss by Al m = 0.5 kg mH2OcH2O(TiH2O – Tf(Al+H2O)) = - mAlcAl(Tf(Al+H2O) – TiAl) -1 -1 c = 880 Jkg K mH2OcH2O(TiH2O – Tf(Al+H2O)) = mAlcAl(TiAl – Tf(Al+H2O)) o Ti = 100 C 1.0 x 4180 x (Tf – 20) 0.5 x 880 x (100 – Tf)

Tf = Tf(Al + H2O) 4,180Tf – 83,600 = 44,000 – 440Tf

(4,180 + 440)Tf = 44,000 + 83,600

H2O 4620Tf = 127,600 m = 1.0 kg 127,600 T = f 4,620 -1 -1 o c = 4180 Jkg K Tf = 27.62 C o Ti = 20 C

Tf = Tf(Al + H2O) The final temperature of the combined aluminium and water is 27.62 oC.

2. A 2.0 kg block of zinc was cooled from a temperature of 80oC by dropping it into 5 kg of water at 25oC. What will be the final temperature of this combination assuming that there is no heat loss or heat gain outside of the system, also assume there is no conversion of water into steam?

Zn Heat loss by Zn = Heat gain by H2O

m = 2.0 kg mZncZn(TiZn – Tf(Zn+H2O)) = mH2OcH2O(TiH2O – Tf(Zn+H2O)) -1 -1 c = 388 Jkg K 2.0 x 388 x (80 – Tf) = 5.0 x 4180 x (Tf – 25) o Ti = 80 C 62,080 – 776Tf = 20,900Tf – 522,500

Tf = Tf(Zn + H2O) 62,080 + 522,500 = (20,900 + 776)Tf

H2O 584,580 = 21,676Tf m = 5.0 kg 584,580 T = f 21,676 -1 -1 o c = 4180 Jkg K Tf = 26.97 C o Ti = 25 C

Tf = Tf(Zn + H2O) The final temperature of the combined zinc and water is equal to 26.97 oC. (In this example the Heat Loss is considered on the LHS and the Heat Gain is on the RHS)

We could make a comment about the change in temperature of the zinc from its original temperature in relation to its mass and specific heat, both of which are less than that for water and how the temperature of the water did not change by much.

Lazarov 2006 This example considers only a single phase interaction; the liquid (water) does not change phase nor does the solid (aluminium or zinc) change phase. The only consideration we must take into account is the specific heat of each of the substances.

If there is a change of phase of a particular object or body of a certain mass then we also need to take into account the latent heat of the object or body that undergoes a phase change or indeed phase changes.

The term latent heat refers to the energy (heat) that is involved when a change of phase occurs. If we consider the particle theory (atomic theory) we could explain the change of phase in a body as the changes that occur in regard to the types of bonds displayed between the particles.

Types of Chemical Bonds  Metallic Bonding; The metal atoms are able to “swap” valence electrons with each of the other atoms in the body or object. This gives rise to freely moving or, mobile, valence electrons that surround a fixed lattice of positive ions. Note that there is no ion as such only a temporary arrangement where the electrons from one atom have gone to another atom and the first atom mentioned has received an electron from elsewhere. All metal elements, and their alloys, undergo this type of bonding.  Ionic bonding; This type of bond involves the electrostatic attraction between ions that have opposite residual charges. For instance a sodium atom (Na) loses an electron to become a positively charged sodium ion (Na+), a chlorine atom (Cl) will accept an electron to become a negatively charged ion (Cl-). This lose or gain of electrons is more significant than the swapping of electrons mentioned above, and may be considered more permanent in nature. The combination of sodium and chloride (note the name changes from chlorine to chloride) ions gives rise to the chemical compound known as sodium chloride (common table salt). In general ionic compounds are formed when metal elements react with non-metal elements.  Covalent Bonding; Atoms involved in this type of bonding share the valence electrons so as to achieve a full complement of valence electrons. This means that their outer shells become “full”. There are two important variations to this type of bond namely Covalent Network Bonding and Covalent Molecular Bonding. o Covalent Network; This gives rise to large collections of atoms arranged in a “network” held together by strong covalent bonds. Each atom is linked to every other atom. The result is very hard and brittle substances such as diamond (an allotrope of carbon) and silicon carbide (used as an abrasive) o Covalent Molecular; A collection of atoms, as little in number as two or as large in number as hundreds, are held together by the same strong covalent bonds, however, there tends to be only weak forces of attraction between the molecules. Small molecules exist as gases at room temperature, eg oxygen (O2) and carbon dioxide (CO2) whereas the larger covalent molecular substances such as paraffin wax (candle wax) are solid at room temperature.

This also gives us some insight as to why the various substances have different values for the specific heat capacity. Now back to latent heat. Even though the chemical composition of a substance does not change, the arrangement of its particles is affected by its heat content.

Lazarov 2006 Consider; water that is cold (frozen) is called ice (chemical formula H2O) and its particles (molecules) are held together in fixed positions typical of a solid. Water that is at room temperature is called water (chemical formula H2O) and its particles are able to slide over each other with relative freedom that is typical of liquids. Finally water that is at an elevated temperature adopts a gaseous condition known as steam (chemical formula H2O) where the particles are moving freely and in a random manner.

There is no difference in the chemical properties of the substance known as water however there is a difference in the physical properties. This difference is mainly due to the “activity” of the particles which is directly related to the amount of heat(energy) that they have. The more heat the faster they will move.

In summary we can refer to the schematic representation below.

Heat Heat Water Steam Ice added added (H O(l)) (H O) (H O ) → 2 → 2 2 (s) Particles are able Particles move Particles held in Heat Heat to slide over each very fast in fixed position Lost Lost other random manner. ← ← Change Change Change in Change in Change in in Phase; in Phase; Temperature; Temperature; Temperature; Latent Latent Specific Heat Specific Heat Specific Heat Heat Heat involved involved involved involved involved

The latent heat is the heat which is required to disturb the bonding so that a change in phase occurs without a change in temperature. This is the reason why latent heat is commonly referred to as the “stored” heat required to cause a change in phase or state.

Note: The rate at which heat energy is applied or removed may cause a rapid phase change accompanied by a change in temperature where the phase change is not readily discernible. Therefore if one were to determine the Melting Point or Boling Point of a peculiar material the rate of heat transfer is required to be slow and steady.

Now it would be wise to consider the events that could or would occur when a hot lump of metal is added to a body of water at room temperature.

Lazarov 2006 Consider. A 5 kg lump of aluminium at an unknown temperature is added to a 1kg mass of water at room temperature (taken to be 25oC). The final temperature of both masses is 100oC. (So far this is similar to the previously solved question.) During the course of adding the hot metal to the water 100g of water were converted to steam at 100oC. Calculate the initial temperature of the mass of aluminium.

Heat gained when Heat gained when Heat Lost by Al = + Al water warms up water becomes steam m = 5 kg mAlcAl(Ti – Tf) = mH2OcH2O(Tf – Ti) + mH2OLv(H2O) o 6 Ti = ? C 5x880x(Ti – 100) = 1x4180x(100 – 25) + 0.1x2.258x10 o 5 5 5 Tf = 100 C 4400Ti – 4.400x10 = 3.135x10 + 2.258x10 -1 -1 5 5 5 cAl = 880 Jkg K 4400Ti = 3.135x10 +2.258x10 + 4.400x10 5 H2O 4400Ti = 9.793x10 5 m = 1 kg Ti = 9.793x10 /4400 o o Ti = 25 C Ti = 2225 C o Tf = 100 C -1 -1 cH2O = 4180 Jkg K This calculation indicates that the initial temperature of the Steam aluminium block was at an initial temperature equal to 2225 oC m = 100 g = 0.1 kg -1 Lv = 2258 kJkg = 2.258 Mjkg-1

A quick reference to the physical properties for the element aluminium reveals that its MPt is 660oC and its BPt is 2470oC,therefore we can see it is beneficial to refer to other sources of data to verify our observations and/or calculations.

Redo the calculations assuming that only 10g of water were changed into steam rather than 100g. Does this give a reasonable answer?

Extra for the experts.

Try to construct a question which provides information so that the calculations arrive at an answer of; The initial temperature of the water is equal to 36 oC

Good luck, if you have any queries see your teacher for assistance.

Lazarov 2006