500g of ice at `0°C` is mixed with 1g steam at `100°C`. The final temperature of the mixture is

Options

A. `0°C`
B. `50°C`
C. `40°C`
D. `100°C`

Answer

To solve the problem of mixing 500g of ice at 0°C with 1g of steam at 100°C, we will use the principles of calorimetry, specifically the concepts of latent heat and heat transfer.

### Step-by-Step Solution:

1. **Identify the Masses and Temperatures:**
- Mass of ice (m_ice) = 500g = 0.5 kg
- Mass of steam (m_steam) = 1g = 0.001 kg
- Initial temperature of ice (T_ice) = 0°C
- Initial temperature of steam (T_steam) = 100°C

2. **Calculate the Heat Required to Melt the Ice:**
- The latent heat of fusion (L_f) of ice = 80 calories/g.
- Heat required to melt the ice (Q_ice) = m_ice × L_f = 500g × 80 cal/g = 40,000 calories.

3. **Calculate the Heat Released by the Steam:**
- The latent heat of vaporization (L_v) of steam = 540 calories/g.
- Heat released by the steam when it condenses to water (Q_steam) = m_steam × L_v = 1g × 540 cal/g = 540 calories.

4. **Compare the Heat Values:**
- Heat required to melt all the ice (Q_ice) = 40,000 calories.
- Heat released by the steam (Q_steam) = 540 calories.
- Since Q_steam < Q_ice, not all the ice will melt. The steam will condense completely, and only a small amount of ice will melt.

5. **Determine the Final State:**
- The steam will convert to water at 0°C, releasing 540 calories.
- The ice will absorb this heat, but since it requires 40,000 calories to melt completely, only a small portion of the ice will melt.

6. **Final Temperature of the Mixture:**
- Since the heat released by the steam is less than what is required to melt all the ice, the final temperature of the mixture will remain at 0°C.

### Conclusion:
The final temperature of the mixture is **0°C**.