The variation of the specific heat of a substance is given by the expression $$ C=A+BT^2$$ where \(A\) and \(B\) are constants and \(T\) is the Celsius temperature. Find the difference between the mean specific heat and specific heat at midpoint.

Discussion:

The variation of the specific heat of a substance is given by the expression $$ C=A+BT^2$$ where \(A\) and \(B\) are constants and \(T\) is the Celsius temperature.

Mean specific heat
$$ \bar{C}=\frac{\int C dT}{dT}=\frac{\int_{0}^{T}(A+BT^2)dT}{T}$$ $$= \frac{AT+BT^3/3}{T}$$ $$=A+BT^2/2$$
C(midpoint)$$ = A+B(T/2)^2$$ $$=A+\frac{BT^2}{4}$$
Hence, the difference between mean specific heat and specific heat at midpoint $$= \bar{C}-C(midpoint)$$ $$=A+BT^2/3-(A+BT^2/4)$$ $$=\frac{BT^2}{12}$$