`4 gm` of steam at `100^(@)C` is added to `20 gm` of water at `46^(@)C` in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of vaporisation `= 540 cal//gm`. Specific heat of water `=1 cal//gm-^(@)C`.

Options

A. 18 gm
B. 20 gm
C. 22 gm
D. 24 gm

Explanation

To solve the problem, we need to find the final mass of water in the container after mixing steam and water. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the heat released by the steam when it condenses to water.
The heat released by the steam (Q1) can be calculated using the formula:
\[ Q_1 = m \times L \]
where:
- \( m \) is the mass of steam (4 gm),
- \( L \) is the latent heat of vaporization (540 cal/gm).

Substituting the values:
\[ Q_1 = 4 \, \text{gm} \times 540 \, \text{cal/gm} = 2160 \, \text{cal} \]

### Step 2: Calculate the heat required to raise the temperature of water from 46°C to 100°C.
The heat required (Q2) can be calculated using the formula:
\[ Q_2 = m \times s \times \Delta T \]
where:
- \( m \) is the mass of water (20 gm),
- \( s \) is the specific heat of water (1 cal/gm°C),
- \( \Delta T \) is the change in temperature (100°C - 46°C = 54°C).

Substituting the values:
\[ Q_2 = 20 \, \text{gm} \times 1 \, \text{cal/gm°C} \times 54 \, \text{°C} = 1080 \, \text{cal} \]

### Step 3: Compare the heat released by the steam and the heat required by the water.
From the calculations:
- \( Q_1 = 2160 \, \text{cal} \)
- \( Q_2 = 1080 \, \text{cal} \)

Since \( Q_1 > Q_2 \), it indicates that not all steam will convert to water.

### Step 4: Determine how much steam will convert to water.
Since the heat released by the steam is greater than the heat required to raise the temperature of the water, we can find the amount of steam that will condense:
- The heat released by half of the steam will equal the heat required to raise the temperature of the water:
\[ \text{Heat released by half steam} = \frac{Q_1}{2} = 1080 \, \text{cal} \]

### Step 5: Calculate the final mass of water in the container.
The final mass of water will be the initial mass of water plus the mass of steam that has condensed:
- Initial mass of water = 20 gm
- Mass of steam that condensed = 2 gm (half of 4 gm)

Thus, the final mass of water is:
\[ \text{Final mass of water} = 20 \, \text{gm} + 2 \, \text{gm} = 22 \, \text{gm} \]

### Conclusion
The final mass of water in the container at thermal equilibrium is **22 gm**.

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