A Carnot engine, whose efficiency is $40 \%$, takes in heat from a source maintained at a temperature of…

A Carnot engine, whose efficiency is $40 \%$, takes in heat from a source maintained at a temperature of $500 \mathrm{~K}$ It is desired to have an engine of efficiency $60 \%$. Then, the intake temperature for the same exhaust (sink) temperature must be
  1. efficiency of Carnot engine cannot be made larger than $50 \%$
  2. $1200 \mathrm{~K}$
  3. $750 \mathrm{~K}$
  4. $600 \mathrm{~K}$

Solution

$\frac{40}{100}=\frac{500-T_S}{500}, T_S=300 \mathrm{~K}$ $\frac{600}{100}=\frac{T-300}{T} \Rightarrow T=750 \mathrm{~K}$

Asked in: JEE Main 2012 (Offline)

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