Question

A Carnot refrigerator absorbs heat from water at $0$ and gives it to a room at $27$. When it converts $2kg$ of water at $0$ into ice at $0$, the work done is (Latent heat of fusions of ice $=333\times {10}^{3}J{kg}^{-1}$

• A. $25\times {10}^{3}J$
• B. $82\times {10}^{3}J$
• C. $65.87\times {10}^{3}J$
• D. $92.52\times {10}^{3}J$

Answer 1

Answer: (C)

$65.87\times {10}^{3}J$

Given - Temperature of sink (water) $T_2=0+273=273K$

Temperature of source (room) $T_1=27+273=300K$

Latent heat of fusion of ice $L=333\times {10}^{3}Jk{g}^{-1}$

mass of water $m=2kg$

Heat taken from 2kg of water at $0^{o}C$ to convert it into ice at $0^{o}C$ ,

$Q_2=mL$ .............................eq1

Now , coefficient of performance of Carnot refrigerator is given by ,

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

Hence , $W=\frac{T_1-T_2}{T_2}{Q}_2$

or $W=\frac{300-273}{273}mL$ (Using eq1)

or $W=\left(\frac{300-273}{273}\right)\times 2\times 333\times {10}^3=65.87\times {10}^{3}J$