1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture is

Options

A. `100^@C`
B. `55^@C`
C. `75^@C`
D. `0^@C`

Explanation

To solve the problem of mixing 1 g of ice at 0°C with 1 g of steam at 100°C and finding the final temperature of the mixture, we can follow these steps:

### Step 1: Identify the heat exchanges
When mixing ice and steam, we need to consider the heat gained by the ice to melt and then warm up to 100°C, and the heat lost by the steam as it condenses into water at 100°C.

### Step 2: Calculate heat required to melt the ice
The latent heat of fusion of ice is 80 kcal/kg (or 0.08 kcal/g). To melt 1 g of ice:
\[ Q_{\text{melt}} = m \cdot L_f = 1 \, \text{g} \cdot 80 \, \text{kcal/kg} = 0.08 \, \text{kcal} \]

### Step 3: Calculate heat required to raise the temperature of water from 0°C to 100°C
The specific heat of water is 1 kcal/kg°C (or 1 kcal/g°C). To raise the temperature of 1 g of water from 0°C to 100°C:
\[ Q_{\text{heat water}} = m \cdot c \cdot \Delta T = 1 \, \text{g} \cdot 1 \, \text{kcal/g°C} \cdot (100 - 0) \, °C = 100 \, \text{kcal} \]

### Step 4: Total heat required by the ice
The total heat required by the ice to become water at 100°C is:
\[ Q_{\text{total ice}} = Q_{\text{melt}} + Q_{\text{heat water}} = 0.08 \, \text{kcal} + 100 \, \text{kcal} = 100.08 \, \text{kcal} \]

### Step 5: Calculate heat released by steam when it condenses
The latent heat of vaporization of steam is 540 kcal/kg (or 0.54 kcal/g). When 1 g of steam condenses to water at 100°C:
\[ Q_{\text{condense}} = m \cdot L_v = 1 \, \text{g} \cdot 540 \, \text{kcal/kg} = 0.54 \, \text{kcal} \]

### Step 6: Determine how much steam condenses
Let \( m \) be the mass of steam that condenses. The heat released by the steam is:
\[ Q_{\text{released}} = m \cdot 540 \, \text{kcal/kg} \]
Setting the heat gained by the ice equal to the heat lost by the steam:
\[ Q_{\text{total ice}} = Q_{\text{released}} \]
\[ 100.08 \, \text{kcal} = m \cdot 0.54 \, \text{kcal/g} \]
Solving for \( m \):
\[ m = \frac{100.08 \, \text{kcal}}{0.54 \, \text{kcal/g}} \approx 185.33 \, \text{g} \]

### Step 7: Analyze the results
Since we only have 1 g of steam, all of it will condense, and the heat released will be sufficient to melt the ice and raise the temperature of the resulting water to 100°C.

### Final Conclusion
After thermal equilibrium is achieved, the temperature of the mixture is:
\[ \text{Final Temperature} = 100°C \]