Question

A Carnot engine has an efficiency of $\frac{1}{6}$. When the temperature of the sink is reduced by ${62}^{\circ }\text{C}$, its efficiency is doubled. The temperatures of the source and the sink are respectively

• A. ${62}^{\circ }\text{C},{124}^{\circ }\text{C}$
• B. ${99}^{\circ }\text{C},{37}^{\circ }\text{C}$
• C. ${124}^{\circ }\text{C},{62}^{\circ }\text{C}$
• D. ${37}^{\circ }\text{C},{99}^{\circ }\text{C}$

Answer 1

The correct option is B ${99}^{\circ }\text{C},{37}^{\circ }\text{C}$

As per the question,

$\eta =\frac{1}{6}=1-\frac{T_{sink}}{T_{source}}$

$\Rightarrow \frac{T_{sink}}{T_{source}}=1-\frac{1}{6}=\frac{5}{6}...\left(1\right)$

When the temperature of sink is reduced by ${62}^{\circ }C$,

${\eta }^{\prime }=\frac{2}{6}=1-\frac{T_{sink}-62}{T_{source}}$

$\Rightarrow \frac{T_{sink}-62}{T_{source}}=\frac{2}{3}...\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow T_{source}\times \frac{1}{6}=62$

$\Rightarrow T_{source}=62\times 6=372\text{K}$

Now, $T_{sink}=\frac{5}{6}\times T_{source}=\frac{5}{6}\times 372=310\text{K}$

$T_{source}\text{in}{}^{\circ }C=372-273={99}^{\circ }\text{C}$

$T_{sink}\text{in}{}^{\circ }C=310-273={37}^{\circ }\text{C}$

Hence, option $\left(B\right)$ is correct.

Why this question? Key tips: While calculating the efficiency of the Carnot's engine, we take absolute temperature, i.e. both the temperature are in kelvin $\left(\text{K}\right)$. Key note: When temperature is reduced by certain amount in ${}^{\circ }\text{C}$, then the same amount of temperature is reduced in kelvin too.