An ideal heat engine operates on a Carnot cyc | vedclass

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Question

(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.Here,$T_H = 227 + 273 = 500\, 	ext{K}$ and $T_L = 127 + 273 = 4

Vedclass · Accepted Answer

$1.2 	imes 10^4\, 	ext{cal}$

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(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$.Here,$T_H = 227 + 273 = 500\, 	ext{K}$ and $T_L = 127 + 273 = 400\, 	ext{K}$.$\eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2$.Efficiency is also defined as $\eta = \frac{W}{Q_H}$,where $W$ is the work done and $Q_H$ is the heat absorbed at the higher temperature.Given $Q_H = 6 	imes 10^4\, 	ext{cal}$.$W = \eta 	imes Q_H = 0.2 	imes 6 	imes 10^4\, 	ext{cal} = 1.2 	imes 10^4\, 	ext{cal}$.

An ideal heat engine operates on a Carnot cycle between $227\,^{\circ}C$ and $127\,^{\circ}C$. It absorbs $6 \times 10^4\, \text{cal}$ at the higher temperature. The amount of heat converted into work is equal to:

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