A Carnot engine working between 300, K and 60 | vedclass

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(C) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is

Vedclass · Accepted Answer

$1600$

Vedclass · Answer

(C) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.Given: $T_1 = 600\, K$,$T_2 = 300\, K$,and work output $W = 800\, J$.Efficiency $\eta = 1 - \frac{300}{600} = 1 - 0.5 = 0.5$.Also,efficiency is defined as $\eta = \frac{W}{Q}$,where $Q$ is the heat supplied from the source.Substituting the values: $0.5 = \frac{800}{Q}$.Therefore,$Q = \frac{800}{0.5} = 1600\, J$.

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

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