The Absorption of Heat by Solids and Liquids — Question 8

Question 8

Two heater powers separate heat capacity from steady loss

At the same temperature, a well-mixed sample heats at .200 K/s with input power 100 W and at .500 K/s with input power 160 W. Its heat capacity $C$ is constant; the environmental loss $P_\ell$ is the same in both instantaneous measurements. Use $C\dot T=P_{\mathrm{in}}-P_\ell$.

1. Infer $C$ and $P_\ell$.

2. If sample mass is .400 kg, find its specific heat. Predict its instantaneous cooling slope when the heater is switched off at that temperature, and the power required for a .100 K/s rise.

Question 8 - Solution

Two heater powers separate heat capacity from steady loss

(a) Method and calculation

$$.200C=100-P_\ell,\quad .500C=160-P_\ell.$$$$C=\frac{60}{.300}=200\,\mathrm{J/K},\quad P_\ell=100-.200(200)=60.0\,\mathrm W.$$

(b) Interpretation and further calculation

$$c=C/m=500\,\mathrm{J/(kg\,K)},\quad\dot T_{\mathrm{off}}=-60/200=-.300\,\mathrm{K/s}.$$$$P_{.100}=P_\ell+C(.100)=80.0\,\mathrm W.$$These are local predictions at the measured temperature, since the loss can change as temperature changes.

Physical check

Taking the difference of the two equations removes the unknown loss without pretending it is zero.