A Carnot’s engine operates with an efficiency of 40% with its sink at `27^(@)C.` With an aim to increase the efficiency to 50 %, the amount of increase in temperature of the source is……….K.

Explanation

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

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

### Step 2: Convert Sink Temperature to Kelvin
The sink temperature is given as \(27^{\circ}C\). To convert this to Kelvin:
\[ T_2 = 27 + 273 = 300 \text{ K} \]

### Step 3: Calculate Initial Source Temperature \(T_1\)
The initial efficiency is given as 40%, or \(0.4\). Using the efficiency formula:
\[ 0.4 = 1 - \frac{300}{T_1} \]
Rearranging the equation gives:
\[ \frac{300}{T_1} = 1 - 0.4 = 0.6 \]
Now, solving for \(T_1\):
\[ T_1 = \frac{300}{0.6} = 500 \text{ K} \]

### Step 4: Calculate New Source Temperature \(T_1'\) for 50% Efficiency
Now, we want to increase the efficiency to 50%, or \(0.5\). Using the efficiency formula again:
\[ 0.5 = 1 - \frac{300}{T_1'} \]
Rearranging gives:
\[ \frac{300}{T_1'} = 1 - 0.5 = 0.5 \]
Now, solving for \(T_1'\):
\[ T_1' = \frac{300}{0.5} = 600 \text{ K} \]

### Step 5: Calculate the Increase in Temperature
The increase in temperature of the source is:
\[ \Delta T = T_1' - T_1 = 600 \text{ K} - 500 \text{ K} = 100 \text{ K} \]

### Final Answer
The amount of increase in temperature of the source is \(100 \text{ K}\).

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