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At very low temperatures, the heat capacity of a crystal grows as the cube of its temperature (in…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

At very low temperatures, the heat capacity of a crystal grows as the cube of its temperature (in kelvin). For each of two identical blocks of a crystal, $C = bT^3$ with $b = 4.0 \times 10^{-4}\ \text{J/K}^4$ , between a few kelvin and $30\ \text{K}$ .

How much heat is needed to warm one block from $10\ \text{K}$ to $20\ \text{K}$ ?
One block is at $10\ \text{K}$ and the other at $30\ \text{K}$ . They are pressed together inside an insulated, evacuated chamber and left until their temperatures are equal. What is the final temperature?
How much heat flows from the warmer block to the cooler one?
What is the average heat capacity of one block between $10\ \text{K}$ and $30\ \text{K}$ (the heat to warm it across that range, divided by $30 - 10\ \text{K}$ )? Compare it with $C$ at $20\ \text{K}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. With $C = bT^3$ , the heat between two temperatures is $\displaystyle\int bT^3\,dT = \frac{b}{4}\left(T_2^4 - T_1^4\right)$ . Use it for each step.

Step 1. Warming (a). $Q = \dfrac{b}{4}\left(20^4 - 10^4\right) = 1.0 \times 10^{-4} \times (160000 - 10000)\ \text{J} = 15.0\ \text{J}$ .

Step 2. Sharing heat (b). Heat lost by the warm block equals heat gained by the cold one:

\displaystyle \frac{b}{4}\left(30^4 - T_f^4\right) = \frac{b}{4}\left(T_f^4 - 10^4\right) \quad\Rightarrow\quad T_f^4 = \frac{10000 + 810000}{2} = 410000\ \text{K}^4

so $T_f = 25.30\ \text{K}$ , much closer to the warm block's temperature than the average, $20\ \text{K}$ .

Step 3. The heat that flows (c). $\dfrac{b}{4}\left(810000 - 410000\right) = 40.00\ \text{J}$ .

Step 4. Average heat capacity (d). $\dfrac{(b/4)(810000 - 10000)}{30 - 10}\ \text{J/K} = 4.00\ \text{J/K}$ , while $C(20\ \text{K}) = b \times 20^3 = 3.20\ \text{J/K}$ . The average is larger, because $C$ grows faster than in a straight line.

Answer: (a) $15.0\ \text{J}$ . (b) $25.30\ \text{K}$ . (c) $40.00\ \text{J}$ . (d) $4.00\ \text{J/K}$ , more than $C$ at $20\ \text{K}$ , which is $3.20\ \text{J/K}$ .

Check. Each block ends at $25.30\ \text{K}$ . The cold block gains $\tfrac{b}{4}(410000 - 10000)\ \text{J}$ and the warm block loses $\tfrac{b}{4}(810000 - 410000)\ \text{J}$ : the same $40.00\ \text{J}$ .

Watch out. The usual mixing rule, with the final temperature as a weighted average, assumes heat capacities that do not change with temperature. Here they change a lot.

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