An ideal gas heat engine operates in a Carnot | vedclass

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(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.Gi

Vedclass · Accepted Answer

$1.2$

Vedclass · Answer

(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ and $T_2$ are the absolute temperatures in Kelvin.Given: $T_1 = 227\,^{\circ}C = 227 + 273 = 500\,K$ and $T_2 = 127\,^{\circ}C = 127 + 273 = 400\,K$.Efficiency $\eta = 1 - \frac{400}{500} = 1 - 0.8 = 0.2$.We know that $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat absorbed at the higher temperature.Given $Q_1 = 6\, kcal$.Therefore,$W = \eta 	imes Q_1 = 0.2 	imes 6 = 1.2\, kcal$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$k$	$2k$	$3k$	$2k$	$k$

An ideal gas heat engine operates in a Carnot cycle between $227\,^{\circ}C$ and $127\,^{\circ}C$. It absorbs $6\, kcal$ at the higher temperature. The amount of heat (in $kcal$) converted into work is equal to

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