Question 1: Thermodynamics & KTG

Question

The temperature of food material in
refrigerator is 4$$^\circ$$C and temperature of
environment is 15$$^\circ$$C. If carnot cycle is used
in its working gas, then find its carnot
efficiency.

Accepted Answer

Given, Temperature of food material in refrigerator,
$$T_2=4^{\circ} \mathrm{C}=273+4=277 \mathrm{~K}$$
Temperature of environment, $$T_1=15^{\circ} \mathrm{C}$$
$$=273+15=288 \mathrm{~K}$$
$$	herefore$$ Carnot efficiency, $$\eta=1-\frac{T_2}{T_1}=1-\frac{277}{288}=0.038$$

Answer

0.038

Answer

0.028

Answer

0.053

Answer

0.072

Thermodynamics & KTG

The temperature of food material in refrigerator is 4$^\circC and temperature of environment is 15^\circ$C. If carnot cycle is used in its working gas, then find its carnot efficiency.

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