A Carnot engine working between 400 , K and 8 | vedclass

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Question

(C) Given: Work output $W = 1200 \, J$,Source temperature $T_1 = 800 \, K$,Sink temperature $T_2 = 400 \, K$.The efficiency $\eta$ of a Carnot engine

Vedclass · Accepted Answer

$2400$

Vedclass · Answer

(C) Given: Work output $W = 1200 \, J$,Source temperature $T_1 = 800 \, K$,Sink temperature $T_2 = 400 \, K$.The efficiency $\eta$ 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 energy supplied by the source.Equating the two expressions: $1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Substituting the values: $1 - \frac{400}{800} = \frac{1200}{Q_1}$.$1 - 0.5 = \frac{1200}{Q_1} \Rightarrow 0.5 = \frac{1200}{Q_1}$.$Q_1 = \frac{1200}{0.5} = 2400 \, J$.Thus,the heat energy supplied to the engine per cycle is $2400 \, J$.

$A$ Carnot engine working between $400 \, K$ and $800 \, K$ has a work output of $1200 \, J$ per cycle. What is the amount of heat energy supplied to the engine from the source per cycle in $J$?

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