`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 beat of fusion of ice `= 3.36xx 10^(5)J kg^(-1)` and latent head of vaporization of water `= 2.26 xx 10^(6)J kg^(-1)`

Explanation

To solve the problem of mixing `1 kg` of ice at `0°C` with `1 kg` of steam at `100°C`, we need to analyze the heat transfer that occurs until thermal equilibrium is reached.

### Step-by-Step Solution:

1. **Identify the heat required to convert ice to water at `0°C`:**
- The latent heat of fusion of ice is given as \( L_f = 3.36 \times 10^5 \, \text{J/kg} \).
- For `1 kg` of ice, the heat required to convert it to water at `0°C` is:
\[ Q_1 = m \cdot L_f = 1 \, \text{kg} \cdot 3.36 \times 10^5 \, \text{J/kg} = 3.36 \times 10^5 \, \text{J} \]

2. **Identify the heat required to raise the temperature of water from `0°C` to `100°C`:**
- The specific heat capacity of water is \( c = 4186 \, \text{J/(kg°C)} \).
- The heat required to raise the temperature of `1 kg` of water from `0°C` to `100°C` is:
\[ Q_2 = m \cdot c \cdot \Delta T = 1 \, \text{kg} \cdot 4186 \, \text{J/(kg°C)} \cdot (100 - 0)°C = 4.186 \times 10^5 \, \text{J} \]

3. **Calculate the total heat required to convert `1 kg` of ice at `0°C` to water at `100°C`:**
- The total heat required is:
\[ Q_{\text{total}} = Q_1 + Q_2 = 3.36 \times 10^5 \, \text{J} + 4.186 \times 10^5 \, \text{J} = 7.546 \times 10^5 \, \text{J} \]

4. **Identify the heat released by steam when it condenses to water at `100°C`:**
- The latent heat of vaporization of water is given as \( L_v = 2.26 \times 10^6 \, \text{J/kg} \).
- The maximum heat released by `1 kg` of steam when it condenses to water is:
\[ Q_{\text{released}} = m \cdot L_v = 1 \, \text{kg} \cdot 2.26 \times 10^6 \, \text{J/kg} = 2.26 \times 10^6 \, \text{J} \]

5. **Determine if the heat released by steam is sufficient to convert ice to water:**
- Since \( Q_{\text{released}} = 2.26 \times 10^6 \, \text{J} \) is greater than \( Q_{\text{total}} = 7.546 \times 10^5 \, \text{J} \), the steam can provide enough heat to convert the ice to water.

6. **Calculate the mass of steam that will condense to provide the required heat:**
- Let \( m \) be the mass of steam that condenses. The heat released by this mass of steam is:
\[ Q_{\text{released}} = m \cdot L_v \]
- Setting this equal to the total heat required:
\[ m \cdot 2.26 \times 10^6 = 7.546 \times 10^5 \]
- Solving for \( m \):
\[ m = \frac{7.546 \times 10^5}{2.26 \times 10^6} \approx 0.333 \, \text{kg} \, \text{or} \, 333 \, \text{g} \]

7. **Determine the final composition of the system:**
- The mass of steam remaining after condensation:
\[ \text{Remaining steam} = 1 \, \text{kg} - 0.333 \, \text{kg} = 0.667 \, \text{kg} \, \text{or} \, 667 \, \text{g} \]
- The mass of water formed from the ice:
\[ \text{Water from ice} = 1 \, \text{kg} + 0.333 \, \text{kg} = 1.333 \, \text{kg} \, \text{or} \, 1333 \, \text{g} \]

### Final Composition:
- Water: `1.333 kg` (or `1333 g`)
- Steam: `0.667 kg` (or `667 g`)