A Carnot engine absorbs 750 J of heat energy from a reservoir at `137^(@)C` and rejects 500 J of heat during each cycle then the temperature of sink is

Options

A. `0.25^(@)C`
B. `0.34^(@)C`
C. `0.44^(@)C`
D. `0.54^(@)C`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Heat absorbed (Q1) = 750 J
- Heat rejected (Q2) = 500 J
- Temperature of the hot reservoir (T1) = 137°C

### Step 2: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
Thus,
\[ T1 = 137 + 273.15 = 410.15 \, K \]

### Step 3: Use the Carnot engine efficiency formula
The efficiency of a Carnot engine can be expressed as:
\[ \frac{Q2}{Q1} = \frac{T2}{T1} \]
Where:
- Q1 = Heat absorbed from the hot reservoir
- Q2 = Heat rejected to the cold reservoir
- T1 = Temperature of the hot reservoir
- T2 = Temperature of the cold reservoir (sink)

### Step 4: Rearrange the formula to find T2
We can rearrange the equation to solve for T2:
\[ T2 = \frac{Q2}{Q1} \times T1 \]

### Step 5: Substitute the known values into the equation
Now substituting the known values:
\[ T2 = \frac{500}{750} \times 410.15 \]

### Step 6: Calculate T2
Calculating the fraction:
\[ \frac{500}{750} = \frac{2}{3} \approx 0.6667 \]
Now substituting this into the equation:
\[ T2 = 0.6667 \times 410.15 \approx 273.43 \, K \]

### Step 7: Convert T2 back to Celsius
To convert Kelvin back to Celsius:
\[ T(°C) = T(K) - 273.15 \]
Thus,
\[ T2(°C) = 273.43 - 273.15 \approx 0.28 \, °C \]

### Final Answer
The temperature of the sink (T2) is approximately **0.28°C**.

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