The Absorption of Heat by Solids and Liquids — Question 2

Question 2

Variable specific heat changes the calorimetric inversion

A .500 kg solid has $c(T)=400+2(T-300)$ J/(kg K) for $300\le T\le400$ K. It starts at 300 K and receives 12.0 kJ without losses, phase change or mechanical work.

1. Integrate $Q=m\int c(T)\,dT$ to find its final temperature.

2. Find the error caused by treating $c$ as its initial value. Determine the mean specific heat over the actual temperature interval.

Question 2 - Solution

Variable specific heat changes the calorimetric inversion

(a) Method and calculation

Let $x=T_f-300$.$$12000=.500\int_0^x(400+2u)du=.500(400x+x^2).$$$$x=\frac{-400+\sqrt{400^2+96000}}2\approx52.9822\,\mathrm K,\quad T_f\approx352.982\,\mathrm K.$$The other quadratic root lies outside the specified domain.

(b) Interpretation and further calculation

$$x_{c_0}=\frac{12000}{.500(400)}=60.0\,\mathrm K.$$$$x_{c_0}-x\approx7.0178\,\mathrm K,\quad\overline c=\frac{Q}{mx}=400+x\approx452.982\,\mathrm{J/(kg\,K)}.$$Using the smaller initial capacity overpredicts the rise.

Physical check

The recovered final temperature lies within the coefficient law's stated range.