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. Here is a step-by-step solution:

### Step 1: Calculate the heat released by steam when it condenses to water.
The heat released when steam condenses to water can be calculated using the formula:
\[ Q_1 = m \cdot L_v \]
where:
- \( m = 1 \, \text{g} \) (mass of steam)
- \( L_v = 540 \, \text{cal/g} \) (latent heat of steam)

Calculating \( Q_1 \):
\[ Q_1 = 1 \, \text{g} \cdot 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice into water.
The heat required to melt ice can be calculated using the formula:
\[ Q_2 = m \cdot L_f \]
where:
- \( m = 1 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of ice)

Calculating \( Q_2 \):
\[ Q_2 = 1 \, \text{g} \cdot 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 3: Calculate the heat required to raise the temperature of the melted ice (water) from 0°C to 100°C.
After the ice melts, we need to raise the temperature of the resulting water from 0°C to 100°C. The heat required for this can be calculated using:
\[ Q_3 = m \cdot c \cdot \Delta T \]
where:
- \( m = 1 \, \text{g} \) (mass of water from melted ice)
- \( c = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \)

Calculating \( Q_3 \):
\[ Q_3 = 1 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 100 \, \text{°C} = 100 \, \text{cal} \]

### Step 4: Total heat required by the ice to become water at 100°C.
The total heat required by the ice to melt and then raise its temperature to 100°C is:
\[ Q_{\text{total, ice}} = Q_2 + Q_3 = 80 \, \text{cal} + 100 \, \text{cal} = 180 \, \text{cal} \]

### Step 5: Compare the heat released by steam and the heat required by ice.
The heat released by the steam is 540 cal, and the heat required by the ice is 180 cal. Since the heat released by the steam is greater than the heat required by the ice, we can conclude that not all steam will convert to water.

### Step 6: Calculate the remaining heat after melting the ice and heating the water.
The remaining heat after the ice has melted and the water has been heated to 100°C is:
\[ Q_{\text{remaining}} = Q_1 - Q_{\text{total, ice}} = 540 \, \text{cal} - 180 \, \text{cal} = 360 \, \text{cal} \]

### Step 7: Determine how much water can be converted back to steam using the remaining heat.
The heat required to convert water back to steam is given by:
\[ Q_{\text{steam}} = m \cdot L_v \]
Let \( m_s \) be the mass of water converted to steam:
\[ 360 \, \text{cal} = m_s \cdot 540 \, \text{cal/g} \]
Solving for \( m_s \):
\[ m_s = \frac{360 \, \text{cal}}{540 \, \text{cal/g}} = \frac{2}{3} \, \text{g} \]

### Step 8: Calculate the final state of the system.
Initially, we had 1 g of steam and 1 g of ice. After the process:
- The mass of steam converted to water is \( 1 - \frac{2}{3} = \frac{1}{3} \, \text{g} \) of steam remains.
- The mass of water from the melted ice is 1 g, plus the 1 g of steam that condensed, giving us a total of 2 g of water.
- The final mixture consists of \( \frac{1}{3} \, \text{g} \) of steam and 2 g of water, all at 100°C.

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

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