Example: Heat capacity 1


Understanding the situation

In order to begin top build an understanding of how objects of different temperatures exchange thermal energy, let's consider the simplest possible example: two amounts of water at different temperatures mixing together. Since if we had equal amounts of water at two different temperatures and mixed them together, we would expect the result to be the average of the two temperatures —and we'd be correct. Let's see what happens if we have different amounts of the same material. And we'll first do the calculation with symbols so we get a general result.

Presenting a sample problem

Source: Adapted from R. Thornton
& D. Sokoloff, with permission

You have two beakers of water with different amounts of water at different temperatures. If one beaker holds 100 g of water at 80 oC and the other holds 200 g of water at 20 oC, what will be their temperature of you mix them together? 

Solving this problem

There is not really enough information to answer this. The beakers are also in thermal contact with the water. Their temperatures will change in the process as well. Our system schema should at least include the two amounts of water and the two beakers (and there seems to be a stirring rod in the picture as well!). And the water in the beakers will interact with the air, the beakers with the hands of the scientist, etc., etc., ...

But we should be able to get a pretty good idea of what the answer will be. When you mix two liquids together and stir, they share thermal energy very quickly. The exchange of heat with the beakers will be considerably slower, and air is a very good insulator so the exchange of heat with the air shouldn't matter if we stir the water together and measure its temperature within a few seconds of the mixing. So let's take as our system the two different amounts of water and only consider exchange of heat between them.

The story of this reduced problem is basically this: The two amounts of water are at different temperatures. When they are mixed together, thermal energy flows from the hot liquid into the cool liquid until their temperatures are equal. The foothold physical principles we need are

  • Conservation of energy — whatever energy leaves on block of water flows into the other;
  • Principle of specific heat — the heat is translated into temperature via the heat capacity of the objects.

Translate these into equation by naming variables. Let the 100 g of water be mass $m_1$ and the 200 g of water be mass $m_2$.

Conservation of energy says

$$Q_2 = -Q_1$$

that is, the heat that enters object 2 is the heat that leaves object 1. (From what we know about how heat works, we are assuming that the energy will flow out of the hot object and go into the cold. So if $Q_2$ is positive, then $Q_1$ must be negative.)

The principle of specific heats says the change in an object's temperature is related to its heat capacity: $C = mc$, its mass times its specific heat. So we have the equations:

$$Q_1 = m_1c_1ΔT_1$$

$$Q_2 = m_2c_2ΔT_2$$

Putting these together,

$$m_1c_1ΔT_1 = -m_2c_2ΔT_2$$

As usual, for practice (and for generality) we'll work this through with symbols and only put the numbers in at the end. We are looking at changes in temperature so writing (as is always true for deltas)

$$ΔT = T^f - T^i$$

where i and f stand for "initial" and "final" respectively.

In our particular case, the two objects (blocks of water) start out at different temperatures and end at the same temperature. Let's call both final temperatures $T$. The equation then becomes:

$$m_1c_1ΔT_1 = -m_2c_2ΔT_2$$

$$m_1c_1(T^f_1 - T^i_1) = -m_2c_2(T^f_2 - T^i_2)$$

Since the masses are both water they have the same specific heat, $c_1 = c_2$, and we can cancel them out. (This ignores a small variation of specific heat with temperature.) Putting in values, we get

$$(100 \mathrm{g})(T - 80 \mathrm{^oC}) = - (200 \mathrm{g})(T - 20 \mathrm{^oC})$$ 

Multiplying out (don't forget the minus sign on the right!) gives

$$(100 \mathrm{g})T  - 8000 \;\mathrm{g ^oC} = - (200 \mathrm{g}) T + 4000 \;\mathrm{g ^oC}$$

Bringing all of the terms with the unknown (T) to one side and solving, we get

$$(100 \mathrm{g})T +  (200 \mathrm{g}) T = 8000 \;\mathrm{g ^oC} + 4000 \;\mathrm{g ^oC}$$

$$(100 \mathrm{g})T +  (200 \mathrm{g}) T = 8000 \;\mathrm{g ^oC} + 4000 \;\mathrm{g ^oC}$$

$$(300 \mathrm{g})T = 12000 \;\mathrm{g ^oC} $$

$$T = \frac{12000}{300} \mathrm{^oC} = 40 \mathrm{^oC} $$

Checking our answer: The units come out right — degrees C, as we expect. And the amount looks right too. Since there is twice as much cold water, the change in the temperature of the hot (40-80=-40) should be twice the magnitude of the change of the cold (40-20=20) and it is.

Joe Redish 11/21/14



Article 433
Last Modified: March 7, 2019