(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1 = 800\, K$ (source temperature) and $T_2 = 400\, K$ (

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(D) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1 = 800\, K$ (source temperature) and $T_2 = 400\, K$ (sink temperature).$\eta = 1 - \frac{400}{800} = 1 - 0.5 = 0.5$.Also,efficiency is defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work output and $Q_1$ is the heat supplied.Given $W = 1200\, J$,we have $0.5 = \frac{1200}{Q_1}$.Therefore,$Q_1 = \frac{1200}{0.5} = 2400\, J$.

Class 11 Physics Thermodynamics Heat Engine and Carnot Cycle

Medium

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

JEE MAIN 2021 All JEE MAIN

A
$3200$
B
$1800$
C
$1600$
D
$2400$

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

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