A Carnot's heat engine works between the temp | vedclass

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(A) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $T_1 = 427^{\circ} C = 427 + 273 = 700\,K$ and $T_2 = 27^{

Vedclass · Accepted Answer

$0.417$

Vedclass · Answer

(A) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Given $T_1 = 427^{\circ} C = 427 + 273 = 700\,K$ and $T_2 = 27^{\circ} C = 27 + 273 = 300\,K$.$\eta = 1 - \frac{300}{700} = 1 - \frac{3}{7} = \frac{4}{7}$.Efficiency is also defined as $\eta = \frac{W}{Q_1}$,where $W$ is the work done per second $(1.0\,kW = 1000\,J/s)$ and $Q_1$ is the heat absorbed per second.$Q_1 = \frac{W}{\eta} = \frac{1000\,J/s}{4/7} = 1000 	imes \frac{7}{4} = 1750\,J/s$.To convert $J/s$ (Watts) to $kcal/s$,we use the conversion $1\,kcal = 4184\,J$ (or approximately $4200\,J$).$Q_1 = \frac{1750}{4184} \approx 0.418\,kcal/s$. Given the options,$0.417\,kcal/s$ is the correct value.

$A$ Carnot's heat engine works between the temperatures $427^{\circ} C$ and $27^{\circ} C$. How much heat in $kcal/s$ should it consume per second to deliver mechanical work at the rate of $1.0\,kW$?

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