Question

The specific heat of a substance varies with temperature t(⁰C) as $c = 0.20 + 0.14t + 0.023t^{2}(cal ⥂ / ⥂ gm ⥂ {^\circ}C)$

The heat required to raise the temperature of 2 gm of substance from 5⁰C to 15⁰C will be

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

Answer 1

Answer: (C)

82 calorie

Answer 2

Heat required to raise the temperature of m gm of substance by $dT$ is given as

dQ = mc dT ⇒ $Q = \int_{}^{}{mcdT}$

∴ To raise the temperature of 2 gm of substance from 5°C to 15°C is

$Q = \int_{5}^{15}{2 \times (0.2 + 0.14t + 0.023t^{2})dT}$ $= 2 \times \left\lbrack 0.2t + \frac{0.14t^{2}}{2} + \frac{0.023t^{3}}{3} \right\rbrack_{5}^{15}$ = 82 calorie