The ratio of the efficiencies of two Carnot e | vedclass

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$

Vedclass · Accepted Answer

$4: 5$

Vedclass · Answer

(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.Let $\Delta T = T_1 - T_2$ be the temperature difference,which is constant for both engines.Thus,$\eta = \frac{\Delta T}{T_1}$.Given the ratio of efficiencies $\frac{\eta_A}{\eta_B} = 1.25 = \frac{5}{4}$.Since $\eta_A = \frac{\Delta T}{T_{1A}}$ and $\eta_B = \frac{\Delta T}{T_{1B}}$,we have $\frac{\eta_A}{\eta_B} = \frac{T_{1B}}{T_{1A}} = \frac{5}{4}$.Therefore,the ratio of the absolute temperatures of the sources $T_{1A} : T_{1B} = 4 : 5$.

The ratio of the efficiencies of two Carnot engines $A$ and $B$ is $1.25$. The temperature difference between the source and the sink is the same in both engines. The ratio of the absolute temperatures of the sources of engines $A$ and $B$ is

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