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 finding the resultant temperature when 1 g of steam is mixed with 1 g of ice, we can follow these steps:

### Step 1: Identify the Initial States
- We have 1 g of steam at 100 °C and 1 g of ice at 0 °C.

### Step 2: Calculate the Heat Released by Steam
- When steam condenses to water at 100 °C, it releases heat. The latent heat of vaporization (L_v) for water is approximately 540 cal/g.
- Heat released by 1 g of steam when it condenses:
\[ Q_{steam} = m \cdot L_v = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 3: Calculate the Heat Required to Melt Ice
- Ice at 0 °C needs heat to melt into water at the same temperature. The latent heat of fusion (L_f) for ice is approximately 80 cal/g.
- Heat required to melt 1 g of ice:
\[ Q_{ice} = m \cdot L_f = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 4: Determine Remaining Heat After Melting Ice
- After melting the ice, the heat available from the steam is:
\[ Q_{remaining} = Q_{steam} - Q_{ice} = 540 \, \text{cal} - 80 \, \text{cal} = 460 \, \text{cal} \]

### Step 5: Calculate the Temperature Rise of the Melted Ice
- The melted ice (now water at 0 °C) will absorb the remaining heat. The specific heat of water (c) is approximately 1 cal/g°C.
- Let \( T_f \) be the final temperature of the water formed from the melted ice. The heat absorbed by the water can be expressed as:
\[ Q_{absorbed} = m \cdot c \cdot (T_f - 0) \]
Setting this equal to the remaining heat:
\[ 460 \, \text{cal} = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (T_f - 0) \]
Therefore,
\[ T_f = 460 \, °C \]

### Step 6: Compare with Boiling Point of Water
- The calculated temperature (460 °C) is above the boiling point of water (100 °C). This indicates that not all steam can condense into water at this temperature.
- The maximum temperature the system can reach is 100 °C, as steam will continue to exist in the gaseous state above this temperature.

### Conclusion
- The equilibrium temperature of the mixture when 1 g of steam is mixed with 1 g of ice is **100 °C**.