Question

A carnot engine has 50% efficiency when the sink temperature is 27°C . What is the increase in temperature of the heat source when its efficiency is 60%?

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Answer 1

Efficiency%=1-T2/T1

Where T1 is the source temperature, T2 is the sink temperature

T2 in Kelvin=27°C+273.15=300 K

Case 1: 50%/100%=1-300/T1

T1=600 K or 327 °C

Case 2: 60%/100%=1-300/T1

T1=750 K or 477 °C

477°C-327°C=150°C