Question

A domestic refrigerator is working between the melting ice temperature and the atmosphere at ${17}^{\circ }\text{C}$. The energy which must be supplied to freeze one $\text{kg}$ of water already at $0^{\circ }\text{C}$ is
(Assume the laent heat for freezing be $80\text{kcal}$)

• A. $2.092\times {10}^4\text{J}$
• B. $3.064\times {10}^4\text{J}$
• C. $2.092\times {10}^6\text{J}$
• D. $3.064\times {10}^6\text{J}$

Answer 1

The correct option is A $2.092\times {10}^4\text{J}$

Given:

$T_2=0^{\circ }\text{C}=0+273\text{K}=273\text{K}$

$T_1={17}^{\circ }\text{C}=(17+273)\text{K}=290\text{K}$

Heat required to be taken out to freeze $1\text{kg}$ of water already at $0^{\circ }\text{C}$ is

$Q_2=mL=1\times 80,000\text{cals}$

$\Rightarrow Q_2=80000\times 4.2\text{J}$


Now, using coefficient of performance of refrigeration is given by

$\frac{Q_2}{W}=\frac{T_2}{T_1-T_2}$

$\Rightarrow W=\frac{Q_2(T_1-T_2)}{T_2}$

$\Rightarrow W=\frac{80000\times 4.2(290-273)}{273}$

$\Rightarrow W=2.092\times {10}^4\text{J}$

So, option $\left(a\right)$ is correct.