`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)kg` of water
D. At equilibrium ,mixture contains `1(2)/(3) kg` of steam

Explanation

To solve the problem of mixing 5 g of steam at 100°C with 10 g of ice at 0°C, we will follow these steps:

### Step 1: Calculate the heat required to convert ice at 0°C to water at 0°C.
The heat required to melt ice can be calculated using the formula:
\[ Q_1 = m \cdot L_F \]
where:
- \( m = 10 \, \text{g} \) (mass of ice)
- \( L_F = 80 \, \text{cal/g} \) (latent heat of fusion)

Substituting the values:
\[ Q_1 = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 2: Calculate the heat required to raise the temperature of the melted ice (water) from 0°C to 100°C.
The heat required to raise the temperature is given by:
\[ Q_2 = m \cdot s \cdot \Delta T \]
where:
- \( m = 10 \, \text{g} \) (mass of water)
- \( s = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100°C - 0°C = 100°C \)

Substituting the values:
\[ Q_2 = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100°C = 1000 \, \text{cal} \]

### Step 3: Calculate the total heat required to convert ice at 0°C to water at 100°C.
The total heat required \( Q_{total} \) is:
\[ Q_{total} = Q_1 + Q_2 = 800 \, \text{cal} + 1000 \, \text{cal} = 1800 \, \text{cal} \]

### Step 4: Calculate the heat released by the steam when it condenses to water.
The heat released when steam condenses can be calculated using:
\[ Q_{released} = m' \cdot L_V \]
where:
- \( m' = \text{mass of steam condensed} \)
- \( L_V = 540 \, \text{cal/g} \) (latent heat of vaporization)

Setting the heat released equal to the heat required:
\[ 1800 \, \text{cal} = m' \cdot 540 \, \text{cal/g} \]

Solving for \( m' \):
\[ m' = \frac{1800 \, \text{cal}}{540 \, \text{cal/g}} = \frac{1800}{540} = \frac{10}{3} \, \text{g} \]

### Step 5: Determine the amount of steam remaining and the total amount of water.
Initially, there were 5 g of steam. After condensing \( \frac{10}{3} \, \text{g} \):
\[ \text{Remaining steam} = 5 \, \text{g} - \frac{10}{3} \, \text{g} = \frac{15}{3} - \frac{10}{3} = \frac{5}{3} \, \text{g} \]

The total amount of water after mixing:
\[ \text{Total water} = 10 \, \text{g (from ice)} + \frac{10}{3} \, \text{g (from steam)} = \frac{30}{3} + \frac{10}{3} = \frac{40}{3} \, \text{g} \]

### Final Result:
- Mass of steam remaining: \( \frac{5}{3} \, \text{g} \)
- Total mass of water: \( \frac{40}{3} \, \text{g} \)