A Carnot engine with efficiency 50% takes heat from a source at 600 K, In order to increase the efficiency to 70 %, keeping the temperature of sink same, the new temperature of the source will be.

Options

A. 900 K
B. 300 K
C. 1000 K
D. 360 K

Answer

To solve the problem, we will use the formula for the efficiency of a Carnot engine:

\[ \eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}} \]

where:
- \(\eta\) is the efficiency of the engine,
- \(T_{\text{sink}}\) is the temperature of the sink,
- \(T_{\text{source}}\) is the temperature of the source.

### Step 1: Calculate the Temperature of the Sink
Given that the initial efficiency \(\eta_1\) is 50%, we can express this as:

\[ \eta_1 = \frac{50}{100} = 0.5 \]

Substituting into the efficiency formula:

\[ 0.5 = 1 - \frac{T_{\text{sink}}}{600} \]

Rearranging gives:

\[ \frac{T_{\text{sink}}}{600} = 1 - 0.5 = 0.5 \]

Thus, we can find \(T_{\text{sink}}\):

\[ T_{\text{sink}} = 600 \times 0.5 = 300 \, \text{K} \]

### Step 2: Calculate the New Temperature of the Source for 70% Efficiency
Now we want to find the new temperature of the source \(T_{\text{source,new}}\) when the efficiency \(\eta_2\) is 70%:

\[ \eta_2 = \frac{70}{100} = 0.7 \]

Using the efficiency formula again:

\[ 0.7 = 1 - \frac{300}{T_{\text{source,new}}} \]

Rearranging gives:

\[ \frac{300}{T_{\text{source,new}}} = 1 - 0.7 = 0.3 \]

Now, we can solve for \(T_{\text{source,new}}\):

\[ T_{\text{source,new}} = \frac{300}{0.3} = 1000 \, \text{K} \]

### Conclusion
The new temperature of the source required to achieve an efficiency of 70% while keeping the sink temperature constant at 300 K is:

\[ \boxed{1000 \, \text{K}} \]