Question

If the absolute temperature of the source of a Carnot's engine is changed from $T_1$ to $T_2$, its efficiency increases from $0.25$ to $0.4$. If the temperature of the sink is constant, then the ratio of $T_1$ and $T_2$ is:

• A. $3:5$
• B. $4:5$
• C. $5:6$
• D. $5:8$

Hint:

The relation between source temperature $T_H$ and efficiency $\eta$ for a constant sink temperature is:
$T_H \propto \frac{1}{1 - \eta}$.
Thus, $\frac{T_1}{T_2} = \frac{1 - \eta_2}{1 - \eta_1} = \frac{1 - 0.4}{1 - 0.25} = \frac{0.6}{0.75} = \frac{4}{5}$.
This simple ratio relation avoids working with fractional equations.

Answer 1

The Correct Option is B

Step 1: Write the sink-to-source ratio for both cases.
From $\eta = 1-\frac{T_L}{T_H}$, we get $\frac{T_L}{T_1} = 1-0.25 = 0.75$ and $\frac{T_L}{T_2} = 1-0.4 = 0.6$.
Step 2: Divide one equation by the other to eliminate the sink temperature.
Since $T_L$ is the same constant in both, \[ \frac{T_L/T_1}{T_L/T_2} = \frac{T_2}{T_1} = \frac{0.75}{0.6} \]
Step 3: Invert to get the required ratio. \[ \frac{T_1}{T_2} = \frac{0.6}{0.75} = \frac{4}{5} \]
\[ \boxed{T_1:T_2 = 4:5} \]