A Carnot engine whose sink is at `300K` has an efficiency of `40%`. By how much should the temperature of source be increased so as to increase its efficiency by `50%` of original efficiency.

Options

A. 275 K
B. 325 K
C. 250 K
D. 380 K

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Identify the given values
- Efficiency of the Carnot engine, \( \eta = 40\% = 0.4 \)
- Temperature of the sink, \( T_L = 300 \, K \)

### Step 2: Use the Carnot efficiency formula
The efficiency of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_L}{T_H} \]
Where \( T_H \) is the temperature of the source.

### Step 3: Rearrange the formula to find \( T_H \)
Substituting the known values into the efficiency formula:
\[ 0.4 = 1 - \frac{300}{T_H} \]
Rearranging gives:
\[ \frac{300}{T_H} = 1 - 0.4 = 0.6 \]
Thus,
\[ T_H = \frac{300}{0.6} = 500 \, K \]

### Step 4: Calculate the new efficiency
The problem states that we want to increase the efficiency by 50% of the original efficiency. Therefore, the increase in efficiency is:
\[ \text{Increase} = 0.5 \times 0.4 = 0.2 \]
Thus, the new efficiency \( \eta' \) is:
\[ \eta' = 0.4 + 0.2 = 0.6 \]

### Step 5: Find the new temperature of the source \( T_H' \)
Using the Carnot efficiency formula again for the new efficiency:
\[ \eta' = 1 - \frac{T_L}{T_H'} \]
Substituting the new efficiency:
\[ 0.6 = 1 - \frac{300}{T_H'} \]
Rearranging gives:
\[ \frac{300}{T_H'} = 1 - 0.6 = 0.4 \]
Thus,
\[ T_H' = \frac{300}{0.4} = 750 \, K \]

### Step 6: Calculate the increase in temperature
To find the increase in temperature of the source:
\[ \Delta T = T_H' - T_H = 750 \, K - 500 \, K = 250 \, K \]

### Final Answer
The temperature of the source should be increased by \( 250 \, K \).

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