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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given data
We have:
- Mass of steam (m_steam) = 2 g
- Mass of water (m_water) = 40 g
- Initial temperature of water (T_initial) = 25 °C
- Final temperature of water (T_final) = 54.3 °C

### Step 2: Calculate the temperature change of the water
The change in temperature (ΔT) of the water can be calculated as:
\[ \Delta T = T_{final} - T_{initial} = 54.3 °C - 25 °C = 29.3 °C \]

### Step 3: Calculate the heat gained by the water
The heat gained by the water (Q_gain) can be calculated using the formula:
\[ Q_{gain} = m_{water} \cdot c_{water} \cdot \Delta T \]
Where \(c_{water} = 1 \, \text{cal/g°C}\). Thus:
\[ Q_{gain} = 40 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 29.3 \, \text{°C} = 1172 \, \text{cal} \]

### Step 4: Calculate the heat lost by the steam
When steam condenses, it releases heat. The heat lost by the steam (Q_lost) is the sum of the heat released during condensation and the heat released as the water cools down to the final temperature:
\[ Q_{lost} = m_{steam} \cdot L + m_{steam} \cdot c_{water} \cdot (100 - T_{final}) \]
Where \(L\) is the latent heat of steam and \(c_{water} = 1 \, \text{cal/g°C}\). Therefore:
\[ Q_{lost} = 2 \, \text{g} \cdot L + 2 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (100 - 54.3) \]
Calculating the second term:
\[ Q_{lost} = 2L + 2 \cdot 45.7 = 2L + 91.4 \]

### Step 5: Set heat gained equal to heat lost
According to the principle of calorimetry, heat gained by the water equals heat lost by the steam:
\[ Q_{gain} = Q_{lost} \]
Substituting the values we calculated:
\[ 1172 = 2L + 91.4 \]

### Step 6: Solve for L
Rearranging the equation:
\[ 2L = 1172 - 91.4 \]
\[ 2L = 1080.6 \]
\[ L = \frac{1080.6}{2} = 540.3 \, \text{cal/g} \]

### Conclusion
The latent heat of steam is approximately:
\[ L \approx 540 \, \text{cal/g} \]