An ideal heat engine exhausts heat at 77,^{ci | vedclass

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temp

Vedclass · Accepted Answer

$227$

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature in Kelvin.Given: $\eta = 30\% = 0.3$,$T_L = 77 + 273 = 350\,K$.Substituting the values: $0.3 = 1 - \frac{350}{T_H}$.$\frac{350}{T_H} = 1 - 0.3 = 0.7$.$T_H = \frac{350}{0.7} = 500\,K$.Converting back to Celsius: $T_H(^{\circ}C) = 500 - 273 = 227\,^{\circ}C$.

An ideal heat engine exhausts heat at $77\,^{\circ}C$. To have a $30\%$ efficiency,it must take heat at what temperature in $^{\circ}C$?

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