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Question: The efficiency of a Carnot engine operating with reservoir temperatures of 100<sup>0</sup>C and –23<...

The efficiency of a Carnot engine operating with reservoir temperatures of 1000C and –230C will be –

A

100−23100\frac{100 - 23}{100}

B

100+23100\frac{100 + 23}{100}

C

100+23373\frac{100 + 23}{373}

D

100−23373\frac{100 - 23}{373}

Answer

100+23373\frac{100 + 23}{373}

Explanation

Solution

h = 1–TLTH\frac{T_{L}}{T_{H}}= 1 –273−23273+100\frac{273 - 23}{273 + 100}=273+100−273+23373\frac{273 + 100 - 273 + 23}{373}

= 100+23273\frac{100 + 23}{273}