An ideal Carnot's engine whose efficiency 40% receives heat of 500K. If the efficiency is to be 50% then the temperature of sink will be

Options

A. 600 K
B. 700 K
C. 800 K
D. 900 K

Answer

To solve the problem, we need to find the temperature of the sink (T2) for a Carnot engine with a given efficiency. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understand the Efficiency Formula**:
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
where \( T_1 \) is the temperature of the heat source and \( T_2 \) is the temperature of the heat sink.

2. **Given Values**:
- The efficiency (η) is given as 50% or 0.50.
- The temperature of the heat source (T1) is given as 500 K.

3. **Substituting the Values into the Efficiency Formula**:
We can rearrange the efficiency formula to find T2:
\[ 0.50 = 1 - \frac{T_2}{500} \]

4. **Rearranging the Equation**:
Rearranging the equation to isolate T2:
\[ \frac{T_2}{500} = 1 - 0.50 \]
\[ \frac{T_2}{500} = 0.50 \]

5. **Solving for T2**:
Now, multiply both sides by 500 to find T2:
\[ T_2 = 500 \times 0.50 \]
\[ T_2 = 250 \text{ K} \]

6. **Final Answer**:
The temperature of the sink (T2) is 250 K.