Steam at `100^(@)C` is passed into `20 g` of water at `10^(@)C` when water acquire a temperature of `80^(@)C`, the mass of water present will be
[Take specific heat of water `= 1 cal g^(-1).^(@) C^(-1)` and latent heat of steam ` = 540 cal g^(-1)`]

Options

A. 24 g
B. 31.5 g
C. 42.5 g
D. 22.5g

Answer

To solve the problem step by step, we will calculate the heat lost by the steam and the heat gained by the water, and then use the principle of calorimetry to find the mass of steam that was converted to water.

### Step 1: Calculate the heat lost by the steam

The heat lost by the steam when it condenses into water at 80°C can be expressed as:
\[ Q_1 = m \cdot L_S + m \cdot S_W \cdot \Delta T_W \]
Where:
- \( m \) = mass of steam (in grams)
- \( L_S \) = latent heat of steam = 540 cal/g
- \( S_W \) = specific heat of water = 1 cal/g°C
- \( \Delta T_W \) = change in temperature of water = \( 100°C - 80°C = 20°C \)

Substituting the values:
\[ Q_1 = m \cdot 540 + m \cdot 1 \cdot 20 \]
\[ Q_1 = 540m + 20m = 560m \text{ cal} \]

### Step 2: Calculate the heat gained by the water

The heat gained by the water when it is heated from 10°C to 80°C can be expressed as:
\[ Q_2 = M_W \cdot S_W \cdot \Delta T_W \]
Where:
- \( M_W \) = mass of water = 20 g
- \( S_W \) = specific heat of water = 1 cal/g°C
- \( \Delta T_W \) = change in temperature of water = \( 80°C - 10°C = 70°C \)

Substituting the values:
\[ Q_2 = 20 \cdot 1 \cdot (80 - 10) \]
\[ Q_2 = 20 \cdot 1 \cdot 70 = 1400 \text{ cal} \]

### Step 3: Apply the principle of calorimetry

According to the principle of calorimetry, the heat lost by the steam is equal to the heat gained by the water:
\[ Q_1 = Q_2 \]
Substituting the expressions we found:
\[ 560m = 1400 \]

### Step 4: Solve for the mass of steam

To find \( m \):
\[ m = \frac{1400}{560} = 2.5 \text{ g} \]

### Step 5: Calculate the total mass of water present

The total mass of water present after the steam condenses is the sum of the original mass of water and the mass of steam that has condensed:
\[ \text{Total mass} = 20 \text{ g} + 2.5 \text{ g} = 22.5 \text{ g} \]

### Final Answer

The total mass of water present will be **22.5 g**.

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