Consider an engine that absorbs 130 text{ cal | vedclass

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(C) The heat absorbed per cycle is $Q_H = 130 	ext{ cal}$.The heat rejected per cycle is $Q_C = 30 	ext{ cal}$.The net heat converted to work per cy

Vedclass · Accepted Answer

$627$

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(C) The heat absorbed per cycle is $Q_H = 130 	ext{ cal}$.The heat rejected per cycle is $Q_C = 30 	ext{ cal}$.The net heat converted to work per cycle is $W_{	ext{net}} = (Q_H - Q_C) 	imes 4.2 	ext{ J/cal} = (130 - 30) 	imes 4.2 = 100 	imes 4.2 = 420 	ext{ J}$.The energy consumed to overcome friction per cycle is $W_f = 2 	ext{ J}$.The useful work delivered to the load per cycle is $W_{	ext{load}} = W_{	ext{net}} - W_f = 420 	ext{ J} - 2 	ext{ J} = 418 	ext{ J}$.The engine operates at $90 	ext{ cycles per minute}$, which is $90/60 = 1.5 	ext{ cycles per second}$.The power delivered to the load is $P = W_{	ext{load}} 	imes 	ext{frequency} = 418 	ext{ J} 	imes 1.5 	ext{ s}^{-1} = 627 	ext{ W}$.

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