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

Options

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

Answer

To solve the problem step by step, we will use the formula for the efficiency of a Carnot engine and the given data.

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

### Step 2: Set up the equations for the two efficiencies
1. For the first case, where the efficiency is 50% (0.5):
\[ 0.5 = 1 - \frac{T_2}{T_1} \]
Rearranging gives:
\[ \frac{T_2}{T_1} = 0.5 \]
Thus, we can express \(T_1\) as:
\[ T_1 = \frac{T_2}{0.5} \]

2. For the second case, where the efficiency is 70% (0.7):
\[ 0.7 = 1 - \frac{T_2}{T_1'} \]
Rearranging gives:
\[ \frac{T_2}{T_1'} = 0.3 \]
Thus, we can express \(T_1'\) as:
\[ T_1' = \frac{T_2}{0.3} \]

### Step 3: Substitute the value of \(T_2\)
Given that the temperature of the sink \(T_2\) is 300 K, we can substitute this value into our equations.

1. For the first case:
\[ T_1 = \frac{300}{0.5} = 600 \text{ K} \]

2. For the second case:
\[ T_1' = \frac{300}{0.3} = 1000 \text{ K} \]

### Step 4: Calculate the increase in temperature
To find the increase in temperature, we calculate:
\[ \Delta T = T_1' - T_1 = 1000 \text{ K} - 600 \text{ K} = 400 \text{ K} \]

### Final Answer
The temperature of the source should be increased by **400 K**.

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