# Example: Heat capacity 2

## Understanding the situation

In our previous example (Example: Thermal energy 1) we poured hot water from a beaker into another beaker of cold water. We only considered the water, ignoring the beakers. In this example, we will consider bringing together two objects with different kinds of specific heats, water and a copper pot, and see how this changes the calculation.

## Presenting a sample problem

Suppose we pour 100 g of hot water (80 oC) into a small 200 g copper pot at root temperature (20 oC). What will be the final temperature of the water and pot? The specific heat of water is 4182 J/kg oC copper and the specific heat of copper is 385 J/kg oC.

## Solving this problem

As in our first example, there really isn't enough information to answer unless we make some assumptions as to what objects play the dominant role. The pot appears to have a handle of a different metal. What about that? We are likely to be able to ignore the role of the air (at least for short times) since it is a good insulator and doesn't conduct heat away quickly. Copper is a good conductor so it will share thermal energy with the water quickly. The handle probably is made of a poor conductor (so that it doesn't get too hot and you can hold the pot). Let's ignore the handle and assume the 200 grams only is for the copper, not the handle.

The story of the problem is similar to that for example 1: We have two objects at different temperatures. When they are brought into contact, thermal energy flows from the hot liquid into the cool pot until their temperatures are equal. As before, 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.

The equations will be the same as in Example 1: $Q_1 = -Q_2$ and $Q = mcΔT$, that is, the heat flowing out of the hot object is that same as that flowing into the cold object (since we are ignoring everything else), and the heat entering or leaving each object changes its temperature according to its heat capacity (mass times specific heat).  The resulting equation is the same, namely:

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

As before, we know the initial temperatures of each object and we want the final temperatures to be the same. The difference here is that not only the masses are different, water and copper have different specific heats. Here are the values: (Those are hyphens in the denominator of the units, not minus signs!)

We can change the units of these to grams (instead of kg) by multiplying by 1 = (1 kg)/(1000 g). We know the initial temperatures and our unknown will be the final temperature which will be the same for both. Putting the numbers into the equation we get

$$(100 \mathrm{g})\bigg(4.182 \frac{\mathrm{J}}{\mathrm{g\;^oC}}\bigg)(T - 80 \mathrm{^oC}) = - (200 \mathrm{g})\bigg(0.385 \frac{\mathrm{J}}{\mathrm{g\;^oC}}\bigg)(T - 20 \mathrm{^oC}) =$$

$$(418.2 \;\mathrm{J/^oC}) T - 33456\;\mathrm{J} = - (770\;\mathrm{J/^oC}) T + 1540\;\mathrm{J}$$

$$(418.2 \;\mathrm{J/^oC} + 770\;\mathrm{J/^oC}) T = (33456 + 1540)\;\mathrm{J}$$

Solving for T gives

$$T = \frac{35000 \;\mathrm{J}}{495.3\;\mathrm{J/^oC}} = 70.7\;\mathrm{^oC}$$

The units came out correctly. It's very close to the temperature of the water, despite the fact that there is twice as much mass of copper! The specific heat of copper is more than 10 times smaller than that of water so it takes much less heat to warm up the copper than it would the same amount of water.

Joe Redish 11/21/14

Article 434