Q121 / 180 · JIPMER 2018 · Physics

A Carnot engine absorbs $6 \times 10^5 \mathrm{cal}$ at $227^{\circ} \mathrm{C}$. The work done per cycle by the engine, if its sink is maintained at $127^{\circ} \mathrm{C}$ is

  • A
    $15 \times 10^8 \mathrm{~J}$
  • B
    $15 \times 10^4 \mathrm{~J}$
  • Correct answer
    C
    $5 \times 10^5 \mathrm{~J}$
  • D
    $2 \times 10^4 \mathrm{~J}$

Solution

Here, $\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}$ $\begin{aligned} & \mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K} \\ & \mathrm{~T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\end{aligned}$ Work done/cycle, $\mathrm{W}=$ ? As $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ or $\mathrm{Q}_2=\frac{\mathrm{T}_2}{\mathrm{~T}_1} \times \mathrm{Q}_1=\frac{400}{500} \times 6 \times 10^5$ $=4.8 \times 10^5 \mathrm{cal}$ $\begin{aligned} \text { As, } & \mathrm{W}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \\ \mathrm{~W} & =1.2 \times 10^5 \mathrm{cal}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \\ \mathrm{~W} & =5.04 \times 10^5 \mathrm{~J}\end{aligned}$

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

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