200 cal of heat is given to a heat engine so that it rejects 150 cal of heat, if source temperature is 400 K, then the sink temperature is

Options

A. 300 K
B. 200 K
C. 100 K
D. 50K

Explanation

To solve the problem, we will use the concept of the heat engine and the relationship between the heat absorbed (Q1), the heat rejected (Q2), and the temperatures of the source (T1) and sink (T2).

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Heat absorbed by the engine (Q1) = 200 cal
- Heat rejected by the engine (Q2) = 150 cal
- Temperature of the source (T1) = 400 K

2. **Use the Efficiency Formula**:
The efficiency (η) of a heat engine can be defined as:
\[ \eta = \frac{Q1 - Q2}{Q1} \]
Here, \( Q1 - Q2 \) is the work done by the engine.

3. **Calculate the Work Done**:
\[ W = Q1 - Q2 = 200 \text{ cal} - 150 \text{ cal} = 50 \text{ cal} \]

4. **Use the Carnot Efficiency Relation**:
For a Carnot engine, the efficiency can also be expressed in terms of the temperatures of the source and sink:
\[ \eta = 1 - \frac{T2}{T1} \]
Therefore, we can write:
\[ \frac{W}{Q1} = 1 - \frac{T2}{T1} \]

5. **Substituting Values**:
Substitute the known values into the efficiency equation:
\[ \frac{50}{200} = 1 - \frac{T2}{400} \]
Simplifying gives:
\[ 0.25 = 1 - \frac{T2}{400} \]

6. **Rearranging the Equation**:
Rearranging the equation to solve for \( T2 \):
\[ \frac{T2}{400} = 1 - 0.25 \]
\[ \frac{T2}{400} = 0.75 \]

7. **Calculating T2**:
Multiply both sides by 400:
\[ T2 = 0.75 \times 400 = 300 \text{ K} \]

### Final Answer:
The temperature of the sink (T2) is **300 K**.