Question

Assume that the domestic refrigerator as a reversible engine working between melting point of ice and atmospheric temperature $\left({17}^{o}C\right)$. The energy which must be supplied to freeze $2$kg of water already at $0^o$C is nearly.

• A. $4.2\times {10}^4$J
• B. $9.5\times {10}^3$J
• C. $3.4\times {10}^4$J
• D. $1.8\times {10}^4$J

Answer 1

The correct option is A $4.2\times {10}^4$J

Given
$T_1={17}^{o}C=17+273=290$K
$T_2=0^{o}C=273$K
So, heat required to be taken out of freeze $2$kg of water already $0^o$C is
$Q_2=mL=20\times 80\times {10}^3$cal
$=16\times {10}^4\times 4.2$J
Also coefficient of performance of refrigerator is
$\frac{Q_2}{W}=\frac{T_2}{T_1-T_2}$
$W=Q_2\frac{T_1-T_2}{T_2}$
$=\frac{16\times {10}^4\times 4.2(290-273)}{290}$
$=4.184\times {10}^{4}J\approx .4.2\times {10}^4$J.