Q167 / 180 · KCET 2018 · Physics

A carnot engine takes \( 300 \) calories of heat from a source at \( 500 \mathrm{~K} \) and rejects \( 150 \) calories of
heat to the sink. The temperature of the sink is

  • A
    \( 125 \mathrm{~K} \)
  • Correct answer
    B
    \( 250 \mathrm{~K} \)
  • C
    \( 750 \mathrm{~K} \)
  • D
    \( 1000 \mathrm{~K} \)

Solution

Given, heat taken from source, $Q_{1}=300$ calories; temperature of source $T_{1}=500 \mathrm{~K} ;$ heat rejected to the sink, $Q_{2}=150$ calories. Temperature of sink, $T_{2}=?$ Now, efficiency, $\eta=\frac{Q_{1}-Q_{2}}{Q}$ Also, efficiency, $\eta=\frac{T_{1}-T_{2}}{T_{1}}$ Therefore, $\frac{Q_{1}-Q_{2}}{Q_{1}}=\frac{T_{1}-T_{2}}{T_{1}}$ $\Rightarrow \frac{300-150}{300}=\frac{500-T_{2}}{500}$ $\Rightarrow \frac{150}{300}=1-\frac{T_{2}}{500}$ $\Rightarrow \frac{1}{2}=1-\frac{T_{2}}{500}$ $\Rightarrow \frac{T_{2}}{500}=1-\frac{1}{2}$ $\Rightarrow T_{2}=500 \times \frac{1}{2}=250 \mathrm{~K}$ Thus, temperature of sink $=250 \mathrm{~K}$

Hence, the correct answer is option: 'B'.

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