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Question: Answered & Verified by Expert
An ideal gas heat engine operates in Carnot cycle between $227^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$. It absorbs $6 \times 10^4 \mathrm{cal}$ of heat at higher temperature. Amount of heat converted to work is
PhysicsThermodynamicsJEE Main
Options:
  • A $2.4 \times 10^4 \mathrm{cal}$
  • B $6 \times 10^4 \mathrm{cal}$
  • C $1.2 \times 10^4 \mathrm{cal}$
  • D $4.8 \times 10^4 \mathrm{cal}$
Solution:
2114 Upvotes Verified Answer
The correct answer is: $1.2 \times 10^4 \mathrm{cal}$
$\begin{aligned} \eta & =\frac{T_1-T_2}{T_1}=\frac{W}{Q} \Rightarrow W=\frac{Q\left(T_1-T_2\right)}{T_1} \\ = & \frac{6 \times 10^4[(227+273)-(273+127)]}{(227+273)} \\ & =\frac{6 \times 10^4 \times 100}{500}=1.2 \times 10^4 \mathrm{cal}\end{aligned}$

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