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Question

A carnot's engine is made to work between $${200}^{o}C$$ and $${0}^{o}C$$ first and then between $${0}^{o}C$$ and $${-200}^{o}C$$. The ratio of efficiencies of the engine in the two cases is:


A
1.73:1
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B
1:1.73
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C
1:1
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D
1:2
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Solution

The correct option is B $$1:1.73$$
Case 1:
$$T_1=0°C=(0+273)K=273K$$
$$T_2=200°=(200+273)K=473K$$
$$\eta_{1}=1-\dfrac {T_1}{T_2}=1-\dfrac {273}{473}=0.422$$
Case 2:
$$T_1=-200°C=(-200+273)K=73K$$
$$T_2=0°=(0+273)K=273K$$
$$\eta_{2}=1-\dfrac {T_1}{T_2}=1-\dfrac {73}{273}=0.73$$

Ratio of efficiencies, $$\eta_1:\eta_2=0.42:0.73=1:1.73$$

Physics

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