A Carnot engine takes 3 times 10^6 calories o | vedclass

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(B) Heat taken by the Carnot engine,$Q = 3 	imes 10^6 	ext{ cal} = 4.2 	imes 3 	imes 10^6 	ext{ J} = 12.6 	imes 10^6 	ext{ J}$.Temperature of t

Vedclass · Accepted Answer

$8.4 	imes 10^6 	ext{ J}$

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(B) Heat taken by the Carnot engine,$Q = 3 	imes 10^6 	ext{ cal} = 4.2 	imes 3 	imes 10^6 	ext{ J} = 12.6 	imes 10^6 	ext{ J}$.Temperature of the source,$T_1 = (627 + 273) 	ext{ K} = 900 	ext{ K}$.Temperature of the sink,$T_2 = (27 + 273) 	ext{ K} = 300 	ext{ K}$.Efficiency of a Carnot engine is given by $\eta = \frac{W}{Q} = 1 - \frac{T_2}{T_1}$.Substituting the values: $\frac{W}{12.6 	imes 10^6} = 1 - \frac{300}{900}$.$\frac{W}{12.6 	imes 10^6} = 1 - \frac{1}{3} = \frac{2}{3}$.$W = \frac{2}{3} 	imes 12.6 	imes 10^6 	ext{ J} = 8.4 	imes 10^6 	ext{ J}$.

$A$ Carnot engine takes $3 \times 10^6$ calories of heat from a reservoir at $627^{\circ} C$ and gives it to a sink at $27^{\circ} C$. The work done by the engine is

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