A Carnot engine, whose sink is at 300 K, has an efficiency of 40%. By how much the source temperature should be changed so as to increase the efficiency to 60% ?

Options

A. 250K increase
B. 250 K decrease
C. 325K increase
D. 325 K decrease

Answer

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

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

### Step 2: Set up the initial conditions
From the problem, we know:
- The sink temperature \(T_L = 300 \, K\)
- The initial efficiency \(η_1 = 40\% = 0.4\)

### Step 3: Calculate the initial source temperature \(T_H\)
Using the efficiency formula:
\[ 0.4 = 1 - \frac{300}{T_H} \]
Rearranging gives:
\[ \frac{300}{T_H} = 0.6 \]
Thus,
\[ T_H = \frac{300}{0.6} = 500 \, K \]

### Step 4: Set up the new efficiency condition
We want to increase the efficiency to \(η_2 = 60\% = 0.6\). Using the same efficiency formula:
\[ 0.6 = 1 - \frac{300}{T_H'} \]
where \(T_H'\) is the new source temperature.

### Step 5: Calculate the new source temperature \(T_H'\)
Rearranging the equation gives:
\[ \frac{300}{T_H'} = 0.4 \]
Thus,
\[ T_H' = \frac{300}{0.4} = 750 \, K \]

### Step 6: Calculate the change in temperature
The change in temperature \(\Delta T\) is given by:
\[ \Delta T = T_H' - T_H = 750 \, K - 500 \, K = 250 \, K \]

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

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