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PART - 2 CH - 11 Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 11 Thermodynamics2 Marks
Question
A Carnot engine works between temperatures $T_1 K$ and $T_2 K$. Prove that to increase its efficiency it will be more effective to reduce the sink temperature to the same value than increasing the source temperature.
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Answer

To increase the efficiency of the Carnot engine between tem. $T _1 K$ and $T _2 K$,
$\eta=\left(1-\frac{T_2}{T_1}\right)$
If the temperature of the source $T_1$ is increased by $\Delta T$ then the resulting efficiency will be :
$\begin{aligned}\eta^{\prime} & =1-\frac{T_2}{\left(T_1+\Delta T\right)} \\
& =\frac{T_1+\Delta T-T_2}{T_1+\Delta T}=\frac{T_1-T_2+\Delta T}{T_1+\Delta T}\ldots\ldots (1)\end{aligned}$
The resulting efficiency on subtracting the sink temperature $T _2$ from $\Delta T$,
$\begin{array}{l}\eta^{\prime \prime}=1-\frac{\left(T_2-\Delta T\right)}{T_1} \\\eta^{\prime \prime}=\frac{T_1-T_2+\Delta T}{T_1}\end{array}$
Equation (2) dividing by equation (1)
$\begin{array}{l}\frac{\eta^{\prime \prime}}{\eta^{\prime}}=\frac{T_1+\Delta T }{ T _1}=1+\frac{\Delta T }{ T _1} \\ \frac{\eta^{\prime \prime}}{\eta^{\prime}}>1 \\ \eta^{\prime \prime}>\eta^{\prime}\end{array}$
Instead of increasing the temperature of the source more efficiency is achieved by reducing the temperature of the sink by the same value. It happens.

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