Consider a refrigerator as reversible heat engine working backwards. It works between $0^{\circ} \ma

Question

Consider a refrigerator as reversible heat engine working backwards. It works between $0^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. The power required to remove heat from the refrigerated space at the rate of 50 kilocalorie per minute is (J = 4.2 Joule/calorie)\

Professor PC Thomas Classes · Answer

A refrigerator is a reversible Carnot's engine working backwards.\ For a Carnot's engine\ $\frac{\mathrm{W}}{\mathrm{Q}_{1}}=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~T}_{1}}$\ where $\mathrm{W}=$ work done\ $\mathrm{Q}_{1}=$ heat absorbed\ $\mathrm{T}_{1}=$ temp. of source\ $\mathrm{T}_{2}$ = temp. of sink\ For a refrigerator $\frac{\mathrm{W}}{\mathrm{Q}_{2}}=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~T}_{2}}$\ $\mathrm{T}_{1}=$ temp of room (higher temperature)\ $\mathrm{T}_{2}=$ temperature of the refrigerated space\ (food material)\ $\mathrm{Q}_{2}=$ heat absorbed from the substance\ $\mathrm{W}=$ work done for absorbing the heat in this process\ $\mathrm{Q}_{2}=50 \mathrm{~K} \mathrm{cal}=50 	imes 10^{3} \mathrm{cal}$ $=50 	imes 10^{3} 	imes 4.2 \mathrm{~J}$\ $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K} \quad \mathrm{~T}_{2}=0^{\circ} \mathrm{C}=273$\ $\frac{\mathrm{W}}{50 	imes 10^{3} 	imes 4.2}=\frac{300-273}{273}=\frac{27}{273}$\ $\mathrm{W}=50 	imes 10^{3} 	imes 4.2 	imes \frac{27}{273} \mathrm{~J}$\ Time $\mathrm{t}=1 \mathrm{mt}=60 \mathrm{~s}$\ $	herefore \mathrm{P}=\frac{50 	imes 10^{3} 	imes 4.2 	imes 27}{1000 	imes 60 	imes 273}$ kilowatt $=\mathrm{kW}$\