A carnot engine whose sink is at 300 K has an efficiency of 50. by how much should the temperature of source be increased so as the efficiency becomes 70% ?

Options

A. 100 K
B. 200 K
C. 300 K
D. 400 K

Answer

To solve the problem step by step, we will use the formula for the efficiency of a Carnot engine and manipulate it to find the required increase in the source temperature.

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

### Step 2: Set Up the Initial Conditions
Given:
- Sink temperature \( T_{\text{sink}} = 300 \, \text{K} \)
- Initial efficiency \( \eta_1 = 50\% = 0.5 \)

Using the efficiency formula:
\[ 0.5 = 1 - \frac{300}{T_{\text{source}}} \]

### Step 3: Solve for the Initial Source Temperature
Rearranging the equation:
\[ \frac{300}{T_{\text{source}}} = 1 - 0.5 \]
\[ \frac{300}{T_{\text{source}}} = 0.5 \]
Multiplying both sides by \( T_{\text{source}} \):
\[ 300 = 0.5 \cdot T_{\text{source}} \]
Now, solving for \( T_{\text{source}} \):
\[ T_{\text{source}} = \frac{300}{0.5} = 600 \, \text{K} \]

### Step 4: Set Up the New Conditions
Now we want to find the new source temperature when the efficiency increases to \( \eta_2 = 70\% = 0.7 \):
\[ 0.7 = 1 - \frac{300}{T_{\text{source new}}} \]

### Step 5: Solve for the New Source Temperature
Rearranging the equation:
\[ \frac{300}{T_{\text{source new}}} = 1 - 0.7 \]
\[ \frac{300}{T_{\text{source new}}} = 0.3 \]
Multiplying both sides by \( T_{\text{source new}} \):
\[ 300 = 0.3 \cdot T_{\text{source new}} \]
Now, solving for \( T_{\text{source new}} \):
\[ T_{\text{source new}} = \frac{300}{0.3} = 1000 \, \text{K} \]

### Step 6: Calculate the Increase in Temperature
The increase in the source temperature \( \Delta T \) is given by:
\[ \Delta T = T_{\text{source new}} - T_{\text{source}} = 1000 \, \text{K} - 600 \, \text{K} = 400 \, \text{K} \]

### Final Answer
The temperature of the source should be increased by **400 K** to achieve an efficiency of 70%.

---