An engine operating between 150,^{circ}C and | vedclass

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(A) The temperatures in Kelvin are:$T_H = 150 + 273 = 423\,K$$T_L = 25 + 273 = 298\,K$Heat absorbed from the reservoir,$Q_H = 500\,J$.For a Carnot eng

Vedclass · Accepted Answer

$147.7$

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(A) The temperatures in Kelvin are:$T_H = 150 + 273 = 423\,K$$T_L = 25 + 273 = 298\,K$Heat absorbed from the reservoir,$Q_H = 500\,J$.For a Carnot engine (no frictional losses),the efficiency $\eta$ is given by:$\eta = \frac{W}{Q_H} = \frac{T_H - T_L}{T_H}$$W = Q_H 	imes \left( \frac{T_H - T_L}{T_H} ight)$$W = 500 	imes \left( \frac{423 - 298}{423} ight)$$W = 500 	imes \left( \frac{125}{423} ight) \approx 147.75\,J$.Rounding to the nearest provided option,the work done is $147.7\,J$.

An engine operating between $150\,^{\circ}C$ and $25\,^{\circ}C$ takes $500\,J$ of heat from a higher temperature reservoir. If there are no frictional losses,the work done by the engine is.....$J$.

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