Question

The specific heat of a substance varies with temperature $t{(}^{\circ }C)$ as
$c=0.20+0.14t+0.023t^2\left(cal/g{m}^{\circ }C\right)$ The heat required to raise the temperature of 2 gm of substance from $5^{\circ }Cto{15}^{\circ }C$

• A. 24 calorie
• B. 56 calorie
• C. 82 calorie
• D. 100 calorie

Answer 1

The correct option is C 82 calorie

Heat required to raise the temperature of m gm of substance by dT is given as
$dQ=mcdT\Rightarrow Q=\int mcdT$
$\therefore$ To raise the temperature of 2 gm of substance from $5^{\circ }Cto{15}^{\circ }C$
$Q={\int }_5^{15}2\times (0.2+0.14t+0.023t^2)dT=2\times {[0.2t+\frac{0.14t^2}{2}+\frac{0.023t^3}{3}]}_5^{15}=82calorie$