Question

Find the time taken by a $500$ W heater to raise the temperature of $50$ kg of material of specific heat capacity $960Jk{g}^{-1}{K}^{-1}$ from ${18}^{0}C$ to ${38}^{0}C$. Assume that all the heat energy supplied by a heater is given to the material.

• A. $32$ min.
• B. $42$ min.
• C. $52$ min.
• D. $62$ min.

Answer 1

The correct option is A $32$ min.

Given :

Power of heater = 500 W

Specific heat capacity , c = 960 J/KgK

Initial temperature, $T_1=18°C=(18+273)K=291K$

Final temperature, $T_2=38°C=(38+273)K=411K$

Mass, m = 50 Kg

Solution :

Heat energy supplied, Q = mc $\delta T$

$\Rightarrow Q=50\times 960\times (411-291)$

$\Rightarrow Q=50\times 960\times 20$

$\Rightarrow Q=960000J$

Power , P = $\frac{Q}{t}$

$\Rightarrow t=\frac{Q}{P}$

$\Rightarrow t=\frac{960000}{500}$

$\Rightarrow t=1920seconds$

Therefore, time , t =$\frac{1920}{60}minutes=32minutes$

Hence, The correct option is A.