A Carnot engine working between 300,K and 600 | vedclass

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(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q}$,where $T_1$ is the source temperatu

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$1600$

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(D) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = \frac{T_1 - T_2}{T_1} = \frac{W}{Q}$,where $T_1$ is the source temperature,$T_2$ is the sink temperature,$W$ is the work output,and $Q$ is the heat supplied.Given: $T_1 = 600\,K$,$T_2 = 300\,K$,and $W = 800\,J$.Rearranging the formula to solve for $Q$: $Q = \left( \frac{T_1}{T_1 - T_2} ight) W$.Substituting the values: $Q = \left( \frac{600}{600 - 300} ight) 	imes 800$.$Q = \left( \frac{600}{300} ight) 	imes 800 = 2 	imes 800 = 1600\,J$.Thus,the heat energy supplied per cycle is $1600\,J$.

$A$ Carnot engine working between $300\,K$ and $600\,K$ has a work output of $800\,J$ per cycle. What is the amount of heat energy supplied to the engine from the source per cycle in $J/cycle$?

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