Question

For a Carnot engine, ratio of work done and heat taken from source is $\frac{1}{4}$. If sink temperature is decreased by ${52}^{\circ }\text{C}$ then the ratio becomes $\frac{1}{2}$. Find the temperature of source.

• A. ${152}^{\circ }\text{C}$
• B. ${184}^{\circ }\text{C}$
• C. ${224}^{\circ }\text{C}$
• D. ${208}^{\circ }\text{C}$

Answer 1

The correct option is D ${208}^{\circ }\text{C}$



Given:
$\frac{W}{Q_1}=\frac{1}{4}=1-\frac{T_2}{T_1}$

$\Rightarrow \frac{T_2}{T_1}=\frac{3}{4}........\left(1\right)$

When the temperature of sink is decreased by ${52}^{\circ }\text{C}$, then,

$\frac{W}{Q_1}=\frac{1}{2}=1-\frac{(T_2-{52}^{\circ }\text{C})}{T_1}$

$\Rightarrow \frac{T_2-{52}^{\circ }\text{C}}{T_1}=\frac{1}{2}$

$\Rightarrow \frac{T_2}{T_1}-\frac{{52}^{\circ }\text{C}}{T_1}=\frac{1}{2}$

$\Rightarrow \frac{3}{4}-\frac{{52}^{\circ }\text{C}}{T_1}=\frac{1}{2}$ [from eq.(1)]

$\Rightarrow \frac{{52}^{\circ }\text{C}}{T_1}=\frac{1}{4}$

$\Rightarrow T_1={208}^{\circ }\text{C}$

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