Q. A refrigerator works between temperature of melting ice and room temperature $(17℃)$. The amount of energy in kWh that must be supplied to freeze 1 kg of water at$0℃\ $is

A) 1.4
B) 1.8
C) 0.058
D) 2.5
Explanation

(c) $T_{2} = 0℃ = 273\ K,$
$T_{1} = 17℃ = 17 + 273 = 290\ K$
Coefficient of performance$= \frac{Q_{2}}{W} = \frac{T_{2}}{T_{1} - T_{2}}$
$\frac{80 \times 1000 \times 4.2}{W} = \frac{273}{290 - 273} = \frac{273}{17}$
$\therefore\ \ \ W = \frac{80 \times 1000 \times 4.2 \times 17}{273}J$
$or\ \ \ \ \ \ W = \frac{33.6 \times 17 \times 10^{4}}{273 \times 3.6 \times 10^{5}}\ kWh$
=0.058 kWh

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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