Heat Engine; Carnot Engine and refrigerator

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Question

A Carnot engine operates between two reservoirs of temperature 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is____________.

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Solution

$E f f i c i e n c y = η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{900} = \frac{2}{3} - - - (1)$
$η = \frac{W}{Q_{a b s o r b e d}} = \frac{W}{Q_{H}} - - - - (2)$
$f r o m (1) a n d (2) w e g e t \Rightarrow Q_{H} = \frac{3}{2} W = \frac{3}{2} \times 1200 = 1800 J$
$W = Q_{H} - Q_{L}$
$\Rightarrow Q_{L} = Q_{H} - W = 1800 - 1200 = 600 J$

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Heat Engine; Carnot Engine and refrigerator

Remember concepts with our Masterclasses.

A Carnot engine operates between two reservoirs of temperature 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is____________.

Efficiency = η=1-T2T1=1−300900=23 ---(1)η=WQabsorbed=WQH ----(2)from (1) and (2) we get ⇒ QH=32W=32×1200=1800J W=QH-QL ⇒QL=QH−W=1800-1200=600J

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