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 $$= \bar{C}-C(midpoint)$$ $$=A+BT^2/3-(A+BT^2/4)$$ $$=\frac{BT^2}{12}$$

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