An ideal gas heat engine operates in a Carnot | vedclass

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(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q_1}$.Here,$T_1 = 227^{\circ}C = 227 + 273 = 500 	e

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$1.2$

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(C) The efficiency $\eta$ of a Carnot engine is given by $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q_1}$.Here,$T_1 = 227^{\circ}C = 227 + 273 = 500 	ext{ K}$ and $T_2 = 127^{\circ}C = 127 + 273 = 400 	ext{ K}$.The heat absorbed at the higher temperature is $Q_1 = 6 	imes 10^4 	ext{ cal}$.Using the formula $W = Q_1 \left( \frac{T_1 - T_2}{T_1} ight)$:$W = 6 	imes 10^4 	imes \left( \frac{500 - 400}{500} ight)$$W = 6 	imes 10^4 	imes \left( \frac{100}{500} ight)$$W = 6 	imes 10^4 	imes 0.2 = 1.2 	imes 10^4 	ext{ cal}$.Thus,the amount of heat converted to work is $1.2 	imes 10^4 	ext{ cal}$.

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

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