1 g of ice at `0^@C` is mixed with 1 g of steam at `100^@C`. After thermal equilibrium is achieved, the temperature of the mixture is

Options

A. `100^@C`
B. `55^@C`
C. `75^@C`
D. `0^@C`

Explanation

To solve the problem of mixing 1 g of ice at 0°C with 1 g of steam at 100°C and finding the final temperature of the mixture after thermal equilibrium is achieved, we can follow these steps:

### Step 1: Identify the given data
- Mass of ice (m_ice) = 1 g
- Initial temperature of ice (T_ice) = 0°C
- Mass of steam (m_steam) = 1 g
- Initial temperature of steam (T_steam) = 100°C

### Step 2: Calculate the latent heat required to convert ice to water
The latent heat of fusion of ice is 80 cal/g. Therefore, the energy required to convert 1 g of ice at 0°C to water at 0°C is:
\[ Q_{ice} = m_{ice} \times L_f = 1 \, \text{g} \times 80 \, \text{cal/g} = 80 \, \text{cal} \]

### Step 3: Calculate the latent heat released by steam when it condenses to water
The latent heat of vaporization of steam is 540 cal/g. Therefore, the energy released when 1 g of steam at 100°C condenses to water at 100°C is:
\[ Q_{steam} = m_{steam} \times L_v = 1 \, \text{g} \times 540 \, \text{cal/g} = 540 \, \text{cal} \]

### Step 4: Determine if there is enough energy from steam to convert ice to water
The energy required to convert the ice to water is 80 cal, and the steam can provide 540 cal. Since 540 cal > 80 cal, there is sufficient energy to convert the ice into water.

### Step 5: Calculate the remaining energy after converting ice to water
After converting the ice to water, the energy left from the steam is:
\[ \text{Remaining energy} = Q_{steam} - Q_{ice} = 540 \, \text{cal} - 80 \, \text{cal} = 460 \, \text{cal} \]

### Step 6: Calculate the temperature increase of the resulting water
Now, we have 1 g of water from the melted ice and 1 g of water from the condensed steam. The total mass of water is:
\[ m_{total} = m_{ice} + m_{steam} = 1 \, \text{g} + 1 \, \text{g} = 2 \, \text{g} \]

The specific heat of water is 1 cal/g°C. We can use the remaining energy to find the temperature increase of the water:
\[ Q = m \times c \times \Delta T \]
Where:
- \( Q \) = remaining energy = 460 cal
- \( m \) = total mass of water = 2 g
- \( c \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature

Rearranging the formula to find \( \Delta T \):
\[ \Delta T = \frac{Q}{m \times c} = \frac{460 \, \text{cal}}{2 \, \text{g} \times 1 \, \text{cal/g°C}} = \frac{460}{2} = 230°C \]

### Step 7: Determine the final temperature
Since the initial temperature of the water from the ice was 0°C, the final temperature of the mixture will be:
\[ T_{final} = T_{initial} + \Delta T = 0°C + 230°C = 230°C \]

However, since water cannot exceed 100°C under normal atmospheric conditions, the final equilibrium temperature will be 100°C.

### Final Answer:
The final temperature of the mixture after thermal equilibrium is achieved is **100°C**.

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