Question

A Carnot freezer takes heat from water at $0^{\circ}C$ inside it and rejects it to the room at a temperature of $27^{\circ}C$. The latent heat of ice is $336 \times 10^3 J \, kg^{-1}$. If $5\, kg$ of water at $0^{\circ}C$ is converted into ice at $0^{\circ}C$ by the freezer, then the energy consumed by the freezer is close to :

• A. $1.67 \times 10^5 \, J$
• B. $1.68 \times 10^6 \, J$
• C. $1.51 \times 10^5 \, J$
• D. $1.71 \times 10^7 \, J$

Answer 1

Answer: (A)

heat required to freeze $5\, kg$
water $=5 \times 336 \times 10^{3}$
$=1680 \times 10^{3}$ Joule
$\Rightarrow Q _{1}=1680 \,KJ$
for carnot's cycle
$\frac{Q_{2}}{Q_{1}}=\frac{T_{2}}{T_{1}}$
$\frac{ Q _{2}}{1680}=\frac{300}{273}$
$Q _{2}=1680 \times \frac{300}{273} \,KJ$
$W=Q_{2}-Q_{1}$
$=1680\left(\frac{300}{273}-1\right)$
$=\frac{1680 \times 27}{273} \times 10^{3} \,J$
$=166.15 \times 10^{3} \,J$
$=1.66 \times 10^{5} \,KJ$