A carnot engine whose sink is at 290 K has an efficiency of 30%. By how much the temperature of the source be increased to have its efficiency equal to 50%, keeping sink temperature constant

Options

A. 200 K
B. 250 K
C. 125 K
D. 300 K

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the efficiency formula
The efficiency (η) of a Carnot engine is given by the formula:
\[ η = 1 - \frac{T_2}{T_1} \]
where:
- \(T_2\) is the temperature of the sink (in Kelvin),
- \(T_1\) is the temperature of the source (in Kelvin).

### Step 2: Set up the initial conditions
We know:
- The sink temperature \(T_2 = 290 \, K\),
- The initial efficiency \(η = 30\% = 0.3\).

### Step 3: Calculate the initial temperature of the source \(T_1\)
Using the efficiency formula:
\[ 0.3 = 1 - \frac{290}{T_1} \]
Rearranging gives:
\[ \frac{290}{T_1} = 1 - 0.3 = 0.7 \]
Thus,
\[ T_1 = \frac{290}{0.7} \]
Calculating this:
\[ T_1 = 414.2857 \, K \approx 414.3 \, K \]

### Step 4: Set up the new efficiency condition
Now, we want the efficiency to be \(50\% = 0.5\) while keeping the sink temperature constant at \(T_2 = 290 \, K\). The new temperature of the source will be \(T_1' = T_1 + T\), where \(T\) is the increase in temperature.

### Step 5: Calculate the new temperature of the source \(T_1'\)
Using the efficiency formula again:
\[ 0.5 = 1 - \frac{290}{T_1'} \]
Rearranging gives:
\[ \frac{290}{T_1'} = 1 - 0.5 = 0.5 \]
Thus,
\[ T_1' = \frac{290}{0.5} = 580 \, K \]

### Step 6: Find the increase in temperature \(T\)
We know:
\[ T_1' = T_1 + T \]
Substituting the known values:
\[ 580 = 414.3 + T \]
Solving for \(T\):
\[ T = 580 - 414.3 = 165.7 \, K \]

### Final Answer
The increase in the temperature of the source is approximately:
\[ \boxed{165.7 \, K} \]