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:
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:
Show 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.
$T_H \propto \frac{1}{1 - \eta}$.
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.
Updated On: Jul 22, 2026
- $3:5$
- $4:5$
- $5:6$
- $5:8$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We need to find the ratio of the source temperatures $T_1$ and $T_2$ for a Carnot engine when its efficiency increases from $0.25$ to $0.4$, assuming the sink temperature remains constant.
Step 2: Key Formula and Approach:
The efficiency $\eta$ of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_L}{T_H} \] where $T_L$ is the absolute temperature of the sink and $T_H$ is the absolute temperature of the source.
We will write the efficiency equations for both cases and solve for the source temperatures in terms of the constant sink temperature $T_L$.
Step 3: Detailed Explanation:
• Case 1: Source temperature is $T_1$, efficiency $\eta_1 = 0.25$:
\[ \eta_1 = 1 - \frac{T_L}{T_1} \] \[ 0.25 = 1 - \frac{T_L}{T_1} \] \[ \frac{T_L}{T_1} = 1 - 0.25 = 0.75 = \frac{3}{4} \] Rearranging to express $T_1$:
\[ T_1 = \frac{4}{3} T_L \quad \text{--- (Equation 1)} \]
• Case 2: Source temperature is $T_2$, efficiency $\eta_2 = 0.4$:
\[ \eta_2 = 1 - \frac{T_L}{T_2} \] \[ 0.4 = 1 - \frac{T_L}{T_2} \] \[ \frac{T_L}{T_2} = 1 - 0.4 = 0.6 = \frac{3}{5} \] Rearranging to express $T_2$:
\[ T_2 = \frac{5}{3} T_L \quad \text{--- (Equation 2)} \]
• Calculate the ratio of $T_1$ to $T_2$:
Divide Equation 1 by Equation 2:
\[ \frac{T_1}{T_2} = \frac{\frac{4}{3} T_L}{\frac{5}{3} T_L} = \frac{4}{5} \] \[ T_1 : T_2 = 4 : 5 \]
Step 4: Final Answer:
The ratio of $T_1$ and $T_2$ is $4:5$, which corresponds to Option (B).
We need to find the ratio of the source temperatures $T_1$ and $T_2$ for a Carnot engine when its efficiency increases from $0.25$ to $0.4$, assuming the sink temperature remains constant.
Step 2: Key Formula and Approach:
The efficiency $\eta$ of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_L}{T_H} \] where $T_L$ is the absolute temperature of the sink and $T_H$ is the absolute temperature of the source.
We will write the efficiency equations for both cases and solve for the source temperatures in terms of the constant sink temperature $T_L$.
Step 3: Detailed Explanation:
• Case 1: Source temperature is $T_1$, efficiency $\eta_1 = 0.25$:
\[ \eta_1 = 1 - \frac{T_L}{T_1} \] \[ 0.25 = 1 - \frac{T_L}{T_1} \] \[ \frac{T_L}{T_1} = 1 - 0.25 = 0.75 = \frac{3}{4} \] Rearranging to express $T_1$:
\[ T_1 = \frac{4}{3} T_L \quad \text{--- (Equation 1)} \]
• Case 2: Source temperature is $T_2$, efficiency $\eta_2 = 0.4$:
\[ \eta_2 = 1 - \frac{T_L}{T_2} \] \[ 0.4 = 1 - \frac{T_L}{T_2} \] \[ \frac{T_L}{T_2} = 1 - 0.4 = 0.6 = \frac{3}{5} \] Rearranging to express $T_2$:
\[ T_2 = \frac{5}{3} T_L \quad \text{--- (Equation 2)} \]
• Calculate the ratio of $T_1$ to $T_2$:
Divide Equation 1 by Equation 2:
\[ \frac{T_1}{T_2} = \frac{\frac{4}{3} T_L}{\frac{5}{3} T_L} = \frac{4}{5} \] \[ T_1 : T_2 = 4 : 5 \]
Step 4: Final Answer:
The ratio of $T_1$ and $T_2$ is $4:5$, which corresponds to Option (B).
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