`1kg` ice at `0^(@)C` is mixed with `1kg` of steam at `100^(@)C`. What will be the composition of the system when thermal equilibrium is reached ? Latent heat of fusion of ice `= 3.36xx 10^(6)J kg^(-1)` and latent heat of vaporization of water `= 2.26 xx 10^(6)J kg^(-1)`

Options

A. 335 g steam and 1665 g water
B. 400 g steam and 1600 g water
C. 465 g steam and 1335 g water
D. None of these

Answer

To solve the problem, we need to analyze the heat exchange between the ice and the steam until thermal equilibrium is reached. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Initial Conditions
- We have 1 kg of ice at 0 °C.
- We have 1 kg of steam at 100 °C.
- Latent heat of fusion of ice \( L_f = 3.36 \times 10^6 \, \text{J/kg} \)
- Latent heat of vaporization of water \( L_v = 2.26 \times 10^6 \, \text{J/kg} \)

### Step 2: Calculate the Heat Required to Melt the Ice
The heat required to convert 1 kg of ice at 0 °C to water at 0 °C is given by:
\[ Q_1 = m \cdot L_f = 1 \, \text{kg} \cdot 3.36 \times 10^6 \, \text{J/kg} = 3.36 \times 10^6 \, \text{J} \]

### Step 3: Calculate the Heat Released by the Steam
The heat released when steam condenses to water at 100 °C is given by:
\[ Q_2 = m \cdot L_v = m \cdot 2.26 \times 10^6 \, \text{J/kg} \]
where \( m \) is the mass of steam that condenses.

### Step 4: Set Up the Heat Balance Equation
At thermal equilibrium, the heat gained by the ice will equal the heat lost by the steam:
\[ Q_1 + Q_2 = 0 \]
Substituting the values we have:
\[ 3.36 \times 10^6 + m \cdot 2.26 \times 10^6 = 0 \]
This can be rearranged to find \( m \):
\[ m \cdot 2.26 \times 10^6 = 3.36 \times 10^6 \]
\[ m = \frac{3.36 \times 10^6}{2.26 \times 10^6} \approx 1.484 \, \text{kg} \]

### Step 5: Determine the Amount of Steam Condensed
Since we started with 1 kg of steam, we can find the amount of steam that remains after condensation:
\[ \text{Mass of steam remaining} = 1 \, \text{kg} - m \approx 1 \, \text{kg} - 1.484 \, \text{kg} = -0.484 \, \text{kg} \]
This indicates that all the steam has condensed, and we have excess heat.

### Step 6: Calculate the Final Composition
The total mass of water formed is the sum of the water from melted ice and the condensed steam:
- Water from melted ice: \( 1 \, \text{kg} \)
- Water from condensed steam: \( m \approx 1.484 \, \text{kg} \)

Thus, the total mass of water when thermal equilibrium is reached is:
\[ \text{Total water} = 1 + 1.484 = 2.484 \, \text{kg} \]
Since all the steam condenses, we have:
- Mass of water = 2.484 kg
- Mass of steam remaining = 0 kg

### Final Answer
When thermal equilibrium is reached, the composition of the system will be:
- Mass of water = 2.484 kg
- Mass of steam = 0 kg