`5`g of steam at `100^(@)C` is mixed with `10`g of ice at `0^(@)C`. Choose correct alternative /s) :- (Given `s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g`)

Options

A. Equilibrium temperature of mixture is `160^(@)C`
B. Equilibrium temperature of mixture is `100^(@)C`
C. At equilibrium ,mixture contains `13(1)/(3)g` of water
D. At equilibrium ,mixture contains `1(2)/(3) g` of steam

Explanation

To solve the problem of mixing steam and ice, we will calculate the heat exchanges involved and determine the final equilibrium temperature and the amounts of water and steam present at that temperature.

### Step 1: Calculate the heat required to convert ice to water at 0°C
The heat required to convert ice to water is given by the formula:
\[ Q_a = m \cdot L_F \]
Where:
- \(m = 10 \, \text{g}\) (mass of ice)
- \(L_F = 80 \, \text{cal/g}\) (latent heat of fusion)

Calculating \(Q_a\):
\[ Q_a = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 2: Calculate the heat required to raise the temperature of the melted ice (water at 0°C) to 100°C
The heat required to raise the temperature of water is given by:
\[ Q_b = m \cdot s \cdot \Delta T \]
Where:
- \(m = 10 \, \text{g}\) (mass of water)
- \(s = 1 \, \text{cal/g°C}\) (specific heat of water)
- \(\Delta T = 100°C - 0°C = 100°C\)

Calculating \(Q_b\):
\[ Q_b = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100°C = 1000 \, \text{cal} \]

### Step 3: Calculate the heat released by steam when it condenses to water at 100°C
The heat released by the steam when it condenses is given by:
\[ Q_c = m \cdot L_V \]
Where:
- \(m = 5 \, \text{g}\) (mass of steam)
- \(L_V = 540 \, \text{cal/g}\) (latent heat of vaporization)

Calculating \(Q_c\):
\[ Q_c = 5 \, \text{g} \cdot 540 \, \text{cal/g} = 2700 \, \text{cal} \]

### Step 4: Determine the net heat exchange
The total heat absorbed by the ice and the water is:
\[ Q_{total} = Q_a + Q_b = 800 \, \text{cal} + 1000 \, \text{cal} = 1800 \, \text{cal} \]

The heat released by the steam is:
\[ Q_c = 2700 \, \text{cal} \]

### Step 5: Compare heat absorbed and released
Since \(Q_c > Q_{total}\), not all steam will condense. The excess heat that will be used to condense some steam is:
\[ Q_{excess} = Q_c - Q_{total} = 2700 \, \text{cal} - 1800 \, \text{cal} = 900 \, \text{cal} \]

### Step 6: Calculate the mass of steam that condenses
The mass of steam that can condense using the excess heat is given by:
\[ m_{condensed} = \frac{Q_{excess}}{L_V} = \frac{900 \, \text{cal}}{540 \, \text{cal/g}} = \frac{10}{6} \, \text{g} = \frac{5}{3} \, \text{g} \approx 1.67 \, \text{g} \]

### Step 7: Calculate the final amounts of water and steam
- Total mass of water after condensation:
\[ m_{water} = 10 \, \text{g} + m_{condensed} = 10 \, \text{g} + \frac{5}{3} \, \text{g} = 10 + 1.67 = \frac{30 + 5}{3} = \frac{35}{3} \, \text{g} \approx 11.67 \, \text{g} \]

- Remaining mass of steam:
\[ m_{steam} = 5 \, \text{g} - m_{condensed} = 5 \, \text{g} - \frac{5}{3} \, \text{g} = \frac{15 - 5}{3} = \frac{10}{3} \, \text{g} \approx 3.33 \, \text{g} \]

### Conclusion
1. The equilibrium temperature of the mixture is \(100°C\).
2. The mixture contains approximately \(13 \frac{1}{3} \, \text{g}\) of water.
3. The remaining steam is approximately \(1 \frac{2}{3} \, \text{g}\).