Question

A refrigerator works between temperature of melting ice and room temperatures $\left(17^{\circ} C \right)$. The amount of energy in $kWh$ that must be supplied to freeze $1\, kg$ of water at $0^{\circ} C$ is

• A. 1.4
• B. 1.8
• C. 0.058
• D. 2.5

Answer 1

Answer: (C)

Given, $T_{2}=0^{\circ} C =273\, K ,$
$T_{1}=17^{\circ} C =17+273=290\, K$
$COP =\frac{ Q _{2}}{W}=\frac{T_{2}}{T_{1}-T_{2}}$
$\Rightarrow \frac{80 \times 1000 \times 4.2}{W} =\frac{273}{290-273}=\frac{273}{17}$
$W=\frac{80 \times 1000 \times 4.2 \times 17}{273} J$
$ W=\frac{33.6 \times 17 \times 10^{4}}{273 \times 3.6 \times 10^{5}} kWh$
$=0.058\, kWh$