2gm of steam condenses when passed through 40 gm of water initially at `25^(@)C` . The condensation of steam raises the temperature of water to `54.3^(@)C` . What is the latent heat of steam

Options

A. 540 cal/g
B. 536 cal/g
C. 270 cal/g
D. 480 cal/g

Answer

To find the latent heat of steam, we can use the principle of conservation of energy. The heat lost by the steam when it condenses and cools down will be equal to the heat gained by the water when its temperature increases.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of steam, \( m_s = 2 \, \text{g} \)
- Mass of water, \( m_w = 40 \, \text{g} \)
- Initial temperature of water, \( T_{w_i} = 25^\circ C \)
- Final temperature of water (and condensed steam), \( T_f = 54.3^\circ C \)

2. **Calculate the temperature change of the water:**
\[ \Delta T_w = T_f - T_{w_i} = 54.3^\circ C - 25^\circ C = 29.3^\circ C \]

3. **Calculate the heat gained by the water:**
The heat gained by the water can be calculated using the formula:
\[ Q_w = m_w \cdot c_w \cdot \Delta T_w \]
where \( c_w \) (specific heat capacity of water) is approximately \( 1 \, \text{cal/g}^\circ C \).
\[ Q_w = 40 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot 29.3^\circ C = 1172 \, \text{cal} \]

4. **Calculate the heat lost by the steam:**
The heat lost by the steam consists of two parts:
- The heat released when steam condenses to water:
\[ Q_{condensation} = m_s \cdot L \]
- The heat released when the condensed steam (now water) cools down from \( 100^\circ C \) to \( 54.3^\circ C \):
\[ Q_{cooling} = m_s \cdot c_w \cdot (100^\circ C - T_f) \]
Combining these, we have:
\[ Q_s = m_s \cdot L + m_s \cdot c_w \cdot (100^\circ C - T_f) \]

5. **Substituting the values:**
\[ Q_s = 2 \, \text{g} \cdot L + 2 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot (100^\circ C - 54.3^\circ C) \]
\[ Q_s = 2L + 2 \cdot 1 \cdot 45.7 = 2L + 91.4 \, \text{cal} \]

6. **Setting the heat gained by water equal to the heat lost by steam:**
\[ Q_w = Q_s \]
\[ 1172 = 2L + 91.4 \]

7. **Solving for \( L \):**
\[ 2L = 1172 - 91.4 \]
\[ 2L = 1080.6 \]
\[ L = \frac{1080.6}{2} = 540.3 \, \text{cal/g} \]

### Final Answer:
The latent heat of steam is approximately \( 540.3 \, \text{cal/g} \).

---