A steam engine intakes 50 g of steam at 100°C per minute and cools it down to 20°C. If latent heat of vaporization of steam is 540 cal `g^(-1)`, then the heat rejected by the steam engine per minute is _______ `xx 10^3` cal. (Given : specific heat capacity of water : 1 cal `g^(-1)`°C-1)

Explanation

To solve the problem, we need to calculate the total heat rejected by the steam engine when it cools down the steam from 100°C to 20°C. The process involves two parts: the heat lost due to condensation of steam and the heat lost due to cooling of water.

### Step 1: Calculate the heat lost due to condensation
The heat lost due to the condensation of steam can be calculated using the formula:
\[ Q_1 = m \cdot L \]
where:
- \( m = 50 \, \text{g} \) (mass of steam)
- \( L = 540 \, \text{cal/g} \) (latent heat of vaporization)

Substituting the values:
\[ Q_1 = 50 \, \text{g} \cdot 540 \, \text{cal/g} = 27000 \, \text{cal} \]

### Step 2: Calculate the heat lost due to cooling of water
Next, we need to calculate the heat lost when the water cools from 100°C to 20°C. This can be calculated using the formula:
\[ Q_2 = m \cdot s \cdot \Delta T \]
where:
- \( s = 1 \, \text{cal/g°C} \) (specific heat capacity of water)
- \( \Delta T = 100°C - 20°C = 80°C \)

Substituting the values:
\[ Q_2 = 50 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 80°C = 4000 \, \text{cal} \]

### Step 3: Calculate the total heat rejected
Now, we can find the total heat rejected by adding the two quantities calculated:
\[ Q_{\text{total}} = Q_1 + Q_2 \]
\[ Q_{\text{total}} = 27000 \, \text{cal} + 4000 \, \text{cal} = 31000 \, \text{cal} \]

### Step 4: Convert to the required format
To express the answer in the required format, we convert calories to kilocalories:
\[ Q_{\text{total}} = 31 \times 10^3 \, \text{cal} \]

Thus, the heat rejected by the steam engine per minute is:
\[ \boxed{31} \]