Question

A refrigerator freezes $5kg$ of water at $0^{o}C$ into ice at $0^{o}C$ in a time interval of $20$ minutes. Assuming that room temperature is ${20}^{o}C$. Calculate the minimum power needed to accomplish it?

• A. $24.4Watt$
• B. $0.1025Watt$
• C. $0.0244Watt$
• D. $102.5Watt$

Answer 1

The correct option is D $102.5Watt$

It is given that

$T_1={20}^{0}C=293K$

$T_2=0^{0}C=273K$

$t=20minutes=1200\sec$

For a refrigerator,

$\frac{Q_2}{W}=\frac{T_2}{T_1-T_2}..........\left(1\right)$

$Q_2=mLL=latentheat=80Kcal/Kg$

$Q_2=5\times 80=400K/cal$

Put all values in (1)

$\frac{400}{W}=\frac{273}{293-273}$

$W=\frac{400\times 20}{273}$

$W=29.3Kcal$

$1Kcal=4.2\times {10}^{3}J$

$W=123.06\times {10}^{3}J$

$power=\frac{W}{t}$

$=\frac{123.06\times {10}^3}{1200}$

$=102.55Watt$