A carnot engine has efficiency `1//5` . Efficiency becomes `1//3` when temperature of sink is decreased by 50 K What is the temperature of sink ?

Options

A. 325 K
B. 375 K
C. 300 K
D. 350 K

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the efficiency of a Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ η = 1 - \frac{T_2}{T_1} \]
where \(T_1\) is the temperature of the hot reservoir (source) and \(T_2\) is the temperature of the cold reservoir (sink).

### Step 2: Set up the first equation using the initial efficiency
Given that the initial efficiency is \( \frac{1}{5} \):
\[ \frac{1}{5} = 1 - \frac{T_2}{T_1} \]
Rearranging this gives:
\[ \frac{T_2}{T_1} = 1 - \frac{1}{5} = \frac{4}{5} \]
This can be expressed as:
\[ T_2 = \frac{4}{5} T_1 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation using the new efficiency
When the temperature of the sink is decreased by 50 K, the new efficiency becomes \( \frac{1}{3} \):
\[ \frac{1}{3} = 1 - \frac{T_2 - 50}{T_1} \]
Rearranging gives:
\[ \frac{T_2 - 50}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
This can be expressed as:
\[ T_2 - 50 = \frac{2}{3} T_1 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 1 into Equation 2
From Equation 1, we have \(T_2 = \frac{4}{5} T_1\). Substitute this into Equation 2:
\[ \frac{4}{5} T_1 - 50 = \frac{2}{3} T_1 \]

### Step 5: Solve for \(T_1\)
To eliminate the fractions, multiply through by 15 (the least common multiple of 5 and 3):
\[ 15\left(\frac{4}{5} T_1 - 50\right) = 15\left(\frac{2}{3} T_1\right) \]
This simplifies to:
\[ 12 T_1 - 750 = 10 T_1 \]
Rearranging gives:
\[ 12 T_1 - 10 T_1 = 750 \]
\[ 2 T_1 = 750 \]
\[ T_1 = 375 \, \text{K} \]

### Step 6: Find \(T_2\)
Using Equation 1 to find \(T_2\):
\[ T_2 = \frac{4}{5} T_1 = \frac{4}{5} \times 375 = 300 \, \text{K} \]

### Final Answer
The temperature of the sink \(T_2\) is **300 K**.

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