if 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is

Options

A. `270^(@)C`
B. `230^(@)C`
C. `100^(@)C`
D. `50^(@)C`

Explanation

To solve the problem of mixing 1 gram of steam with 1 gram of ice and finding the resultant temperature of the mixture, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial States**:
- We have 1 gram of steam at 100°C.
- We have 1 gram of ice at 0°C.

2. **Latent Heat Values**:
- The latent heat of fusion (ice to water) is 80 calories/gram.
- The latent heat of vaporization (steam to water) is 536 calories/gram.

3. **Energy Required to Melt Ice**:
- To convert 1 gram of ice at 0°C to water at 0°C, we need:
\[ Q_{\text{ice}} = m \cdot L_f = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

4. **Energy Released by Steam**:
- To convert 1 gram of steam at 100°C to water at 100°C, the energy released is:
\[ Q_{\text{steam}} = m \cdot L_v = 1 \, \text{g} \cdot 536 \, \text{cal/g} = 536 \, \text{cal} \]

5. **Comparison of Energy**:
- The energy required to melt the ice (80 cal) is much less than the energy released by the steam (536 cal). This means that all the ice will melt, and there will still be some energy left from the steam.

6. **Final State of the Mixture**:
- After the ice melts, we have 1 gram of water at 0°C.
- The steam will release enough energy to convert itself into water, but since it has excess energy (536 cal - 80 cal = 456 cal), the remaining energy will heat the water formed from the ice.

7. **Heating the Water from Ice**:
- The 1 gram of water from the melted ice will absorb some of the remaining energy. The temperature increase can be calculated using:
\[ Q = m \cdot s \cdot \Delta T \]
- Where \( s \) (specific heat of water) is approximately 1 cal/g°C.
- The energy available for heating the water is 456 cal.
- Setting up the equation:
\[ 456 \, \text{cal} = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot \Delta T \]
- Solving for \( \Delta T \):
\[ \Delta T = 456 \, \text{°C} \]

8. **Final Temperature Calculation**:
- The final temperature of the mixture will be:
\[ T_{\text{final}} = 0°C + 456°C = 100°C \]

### Conclusion:
The resultant temperature of the mixture when 1 gram of steam is mixed with 1 gram of ice is **100°C**.