A solid copper sphere of radius 0.010 m is painted black (treat it as a black body)…

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

A solid copper sphere of radius $0.010\ \text{m}$ is painted black (treat it as a black body). It hangs on a thin thread in an evacuated enclosure whose walls are kept at $300\ \text{K}$ . A tiny heater inside the sphere keeps it at $600\ \text{K}$ . Copper has density $8900\ \text{kg/m}^3$ and specific heat $385\ \text{J/(kg K)}$ ; take $\sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2\,\text{K}^4)$ . Ignore heat flow along the thread, and take the sphere's temperature to be the same throughout.

What is the power of the heater?
The heater is switched off. At what rate does the sphere's temperature start to fall?
How long does the sphere take to cool from $600\ \text{K}$ to $500\ \text{K}$ ? (You may use $\displaystyle\int \frac{dT}{T^4 - a^4} = \frac{1}{4a^3}\left[\ln\frac{T - a}{T + a} - 2\tan^{-1}\frac{T}{a}\right]$ .)
How long would a sphere of the same copper, but of twice the radius, take for the same cooling?

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. Balance the net radiation against the heater, then against the loss of stored heat. Integrate for the time.

Step 1. The sphere. Its surface area is $4\pi r^2 = 12.566\ \text{cm}^2$ , its mass $37.28\ \text{g}$ , and its heat capacity $mc = 14.35\ \text{J/K}$ .

Step 2. The heater (a). The sphere radiates $\sigma A T^4$ and takes in $\sigma A T_0^4$ from the walls. The heater makes up the difference:

\displaystyle P = \sigma A\left(T^4 - T_0^4\right) = 5.67 \times 10^{-8} \times 1.2566 \times 10^{-3} \times 1.2150 \times 10^{11}\ \text{W} = 8.66\ \text{W}

Step 3. The first rate (b). $\dfrac{dT}{dt} = -\dfrac{P}{mc} = -\dfrac{8.66}{14.35}\ \text{K/s} = -0.603\ \text{K/s}$ .

Step 4. The time (c). Separate the variables in $mc\,\dfrac{dT}{dt} = -\sigma A(T^4 - T_0^4)$ :

\displaystyle t = \frac{mc}{\sigma A}\int_{500}^{600} \frac{dT}{T^4 - T_0^4}

Here $\dfrac{mc}{\sigma A} = 2.0144 \times 10^{11}\ \text{K}^4\,\text{s}$ , and with the given formula ( $a = 300\ \text{K}$ ) the integral is $1.2420 \times 10^{-9}\ \text{K}^{-3}$ . So $t = 250\ \text{s}$ .

Step 5. A larger sphere (d). The heat capacity grows as $r^3$ and the area as $r^2$ , so $mc/\sigma A$ grows as $r$ . The same cooling takes twice as long: $500\ \text{s}$ .

Answer: (a) $8.66\ \text{W}$ . (b) $0.603\ \text{K/s}$ . (c) $250\ \text{s}$ . (d) $500\ \text{s}$ .

Check. The rate falls from $0.603\ \text{K/s}$ at $600\ \text{K}$ to $0.270\ \text{K/s}$ at $500\ \text{K}$ . Cooling the whole $100\ \text{K}$ at the first rate would take $166\ \text{s}$ ; at the last rate it would take much longer. The answer, $250\ \text{s}$ , lies between, nearer the start because the rate is high there.

Watch out. Newton's law of cooling (a rate proportional to $T - T_0$ ) only works for small temperature differences. Here the differences are large, so the full $T^4$ law is needed.

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