A heat engine operating between temperatures | vedclass

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Question

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta_1 = \frac{1}{6}$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}

Vedclass · Accepted Answer

$372 \ K, 310 \ K$

Vedclass · Answer

(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $\eta_1 = \frac{1}{6}$,we have $\frac{1}{6} = 1 - \frac{T_2}{T_1}$,which implies $\frac{T_2}{T_1} = \frac{5}{6}$,or $T_2 = \frac{5}{6} T_1$ (Equation $i$).When $T_2$ is lowered by $62 \ K$,the new temperature is $(T_2 - 62) \ K$. The new efficiency is $\eta_2 = \frac{1}{3}$.Thus,$\frac{1}{3} = 1 - \frac{T_2 - 62}{T_1}$.Rearranging gives $\frac{T_2 - 62}{T_1} = 1 - \frac{1}{3} = \frac{2}{3}$.Substituting $T_2 = \frac{5}{6} T_1$ into this equation:$\frac{\frac{5}{6} T_1 - 62}{T_1} = \frac{2}{3}$.$\frac{5}{6} - \frac{62}{T_1} = \frac{2}{3}$.$\frac{62}{T_1} = \frac{5}{6} - \frac{2}{3} = \frac{5-4}{6} = \frac{1}{6}$.Therefore,$T_1 = 62 	imes 6 = 372 \ K$.Now,$T_2 = \frac{5}{6} 	imes 372 = 310 \ K$.

$A$ heat engine operating between temperatures $T_1$ and $T_2$ has an efficiency of $\frac{1}{6}$. When $T_2$ is lowered by $62 \ K$,its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ respectively are:

Similar Questions

$Assertion :$ The Carnot cycle is useful in understanding the performance of heat engines.
$Reason :$ The Carnot cycle provides a way of determining the maximum possible efficiency achievable with reservoirs of given temperatures.

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An ideal heat engine exhausts heat at $77\,^{\circ}C$. To have a $30\%$ efficiency,it must take heat at what temperature in $^{\circ}C$?

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