A Carnot engine working between 300 K and 40 | vedclass

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(B) Given:Temperature of source,$T_1 = 400 \ K$Temperature of sink,$T_2 = 300 \ K$Useful work done,$W = 800 \ J$Efficiency of a Carnot engine is given

Vedclass · Accepted Answer

$3200$

Vedclass · Answer

(B) Given:Temperature of source,$T_1 = 400 \ K$Temperature of sink,$T_2 = 300 \ K$Useful work done,$W = 800 \ J$Efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Also,efficiency is defined as $\eta = \frac{W}{Q_1}$,where $Q_1$ is the heat supplied.Equating the two,we get $1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Substituting the values: $1 - \frac{300}{400} = \frac{800}{Q_1}$.$\frac{100}{400} = \frac{800}{Q_1}$.$\frac{1}{4} = \frac{800}{Q_1}$.$Q_1 = 800 	imes 4 = 3200 \ J$.Thus,the heat energy supplied to the engine is $3200 \ J$.

$A$ Carnot engine working between $300 \ K$ and $400 \ K$ produces $800 \ J$ of useful work. The amount of heat energy supplied to the engine from the source is: (in $J$)

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