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

Options

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

Answer

To solve the problem, we need to use the formula for the efficiency of a Carnot engine, which is given by:

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

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

### Step 1: Find the initial temperature of the source

Given that the efficiency (\(\eta\)) is 50% (or 0.5) and the temperature of the sink (\(T_{\text{sink}}\)) is 300 K, we can rearrange the formula to find \(T_{\text{source}}\):

\[ 0.5 = 1 - \frac{300}{T_{\text{source}}} \]

Rearranging gives:

\[ \frac{300}{T_{\text{source}}} = 1 - 0.5 = 0.5 \]

Multiplying both sides by \(T_{\text{source}}\):

\[ 300 = 0.5 \cdot T_{\text{source}} \]

Now, solving for \(T_{\text{source}}\):

\[ T_{\text{source}} = \frac{300}{0.5} = 600 \text{ K} \]

### Step 2: Find the new temperature of the source for 70% efficiency

Now, we want to find the new temperature of the source when the efficiency is 70% (or 0.7). Using the same formula:

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

Rearranging gives:

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

Multiplying both sides by \(T_{\text{source,new}}\):

\[ 300 = 0.3 \cdot T_{\text{source,new}} \]

Now, solving for \(T_{\text{source,new}}\):

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

### Step 3: Calculate the increase in temperature of the source

Now, we can find the increase in temperature of the source:

\[ \Delta T = T_{\text{source,new}} - T_{\text{source}} \]

Substituting the values we found:

\[ \Delta T = 1000 \text{ K} - 600 \text{ K} = 400 \text{ K} \]

### Final Answer

The temperature of the source should be increased by **400 K** to achieve an efficiency of 70%.

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