The efficiency of Carnot's enegine is `50%`. The temperature of its sink is `7^(@)C`. To increase its efficiency to `70%`. What is the increase in temperature of the source?

Explanation

To solve the problem of finding the increase in temperature of the source when the efficiency of a Carnot engine changes from 50% to 70%, we will follow these steps:

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

### Step 2: Convert Sink Temperature to Kelvin
The temperature of the sink is given as \( 7^\circ C \). To convert this to Kelvin:
\[ T_2 = 7 + 273 = 280 \, K \]

### Step 3: Calculate Source Temperature for 50% Efficiency
For the first case where the efficiency is \( 50\% \) (or \( 0.5 \)):
\[ 0.5 = 1 - \frac{280}{T_1} \]
Rearranging gives:
\[ \frac{280}{T_1} = 0.5 \implies T_1 = \frac{280}{0.5} = 560 \, K \]

### Step 4: Calculate Source Temperature for 70% Efficiency
Now, for the second case where the efficiency is \( 70\% \) (or \( 0.7 \)):
\[ 0.7 = 1 - \frac{280}{T_1'} \]
Rearranging gives:
\[ \frac{280}{T_1'} = 0.3 \implies T_1' = \frac{280}{0.3} = \frac{2800}{3} \approx 933.33 \, K \]

### Step 5: Calculate the Increase in Source Temperature
The increase in temperature of the source (\( \Delta T_1 \)) is given by:
\[ \Delta T_1 = T_1' - T_1 \]
Substituting the values we found:
\[ \Delta T_1 = 933.33 \, K - 560 \, K = 373.33 \, K \]

### Conclusion
The increase in temperature of the source required to change the efficiency from 50% to 70% is approximately \( 373.33 \, K \).

---