1 g of steam at `100^@C` and an equal mass of ice at `0^@C` are mixed. The temperature of the mixture in steady state will be (latent heat of steam`=540 cal//g`, latent heat of ice `=80 cal//g`,specific heat of water `=1 cal//g^@C`)

Options

A. `50^@C`
B. `100^@C`
C. `67^@C`
D. None of these

Explanation

To solve the problem of mixing 1 g of steam at 100°C with 1 g of ice at 0°C, we need to analyze the heat transfer involved in the phase changes and temperature changes.

### Step-by-Step Solution:

1. **Identify the Heat Transfer Processes**:
- The steam will condense into water at 100°C, releasing heat.
- The ice will melt into water at 0°C, absorbing heat.
- The resulting water from the steam will cool down to the final temperature of the mixture.
- The resulting water from the melted ice will warm up to the final temperature of the mixture.

2. **Calculate the Heat Released by Steam**:
- When 1 g of steam condenses to water at 100°C, it releases heat equal to the latent heat of steam:
\[ Q_{\text{steam}} = m \cdot L_{\text{steam}} = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

3. **Calculate the Heat Required to Melt Ice**:
- When 1 g of ice melts to water at 0°C, it absorbs heat equal to the latent heat of ice:
\[ Q_{\text{ice}} = m \cdot L_{\text{ice}} = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

4. **Determine the Heat Balance**:
- The heat released by the steam (540 cal) is greater than the heat required to melt the ice (80 cal). Therefore, after melting the ice, there will still be excess heat available from the steam.

5. **Calculate the Remaining Heat After Melting Ice**:
- After melting the ice, the heat remaining from the steam is:
\[ Q_{\text{remaining}} = Q_{\text{steam}} - Q_{\text{ice}} = 540 \, \text{cal} - 80 \, \text{cal} = 460 \, \text{cal} \]

6. **Calculate the Final Temperature of the Mixture**:
- The water produced from the melted ice (1 g at 0°C) will absorb heat from the remaining heat of the steam until they reach thermal equilibrium.
- Let \( T_f \) be the final temperature. The heat absorbed by the melted ice (now water) to reach \( T_f \) is:
\[ Q_{\text{absorbed}} = m \cdot c \cdot (T_f - 0) = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot T_f = T_f \, \text{cal} \]
- The heat released by the steam as it cools down from 100°C to \( T_f \) is:
\[ Q_{\text{released}} = m \cdot c \cdot (100 - T_f) = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (100 - T_f) = 100 - T_f \, \text{cal} \]

7. **Set Up the Equation for Heat Balance**:
- Since the heat absorbed by the melted ice equals the heat released by the steam:
\[ T_f = 100 - T_f + 460 \]
- Rearranging gives:
\[ 2T_f = 560 \implies T_f = 280 \, \text{°C} \]

8. **Conclusion**:
- Since the final temperature cannot exceed 100°C (the boiling point of water), we conclude that the final temperature of the mixture will be 100°C.

### Final Answer:
The temperature of the mixture in steady state will be **100°C**.