Q137 / 160 · AP EAMCET 2024 (19 May Shift 2) · Physics

An ideal heat engine operates in Carnot cycle between $127^{\circ} \mathrm{C}$ and $27^{\circ} \mathrm{C}$. It absorbs $5 \times 10^4 \mathrm{cal}$ of heat at higher temperature. Amount of heat converted to work is

  • A
    $4.8 \times 10^4 \mathrm{cal}$
  • B
    $2.4 \times 10^4 \mathrm{cal}$
  • Correct answer
    C
    $1.25 \times 10^4 \mathrm{cal}$
  • D
    $6 \times 10^4 \mathrm{cal}$

Solution

For carnot cycle, $\begin{aligned} & \mathrm{T}_1=127^{\circ} \mathrm{c}, \mathrm{T}_2=27^{\circ} \mathrm{c}, \mathrm{Q}_1=5 \times 10^4 \mathrm{cal} \\ & \therefore \text { Efficiency, } \eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}=\frac{\mathrm{W}}{\mathrm{Q}_1} \\ & \Rightarrow 1-\frac{27+273}{127+273}=\frac{\mathrm{W}}{5 \times 10^4} \\ & \therefore \mathrm{~W}=\frac{1}{4} \times 5 \times 10^4=1.25 \times 10^4 \mathrm{cal}\end{aligned}$

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

Relevant Questions