A mass m of steam at `100^(@)C` is to passed into a vessel containing 10g of ice and 100g of water at `0^(@)C` so that all the ice is melted and the temperature is raised to `5^(@)C` neglecting heat absorbed by the vessel, we get

Options

A. m = 2.1 g
B. m = 4.2 g
C. m = 6.3g
D. m = 8.4g

Answer

To solve the problem, we need to calculate the mass of steam required to melt the ice and raise the temperature of the resulting water to 5°C. We'll break it down step by step.

### Step 1: Calculate the heat required to melt the ice
The mass of ice is given as 10 g. The latent heat of fusion of ice is 80 cal/g. Therefore, the heat required to melt the ice (Q1) is calculated as:

\[ Q_1 = \text{mass of ice} \times \text{latent heat of fusion} = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 2: Calculate the heat required to raise the temperature of water
After melting, we have 10 g of water from the ice and 100 g of water already present, giving a total mass of water as 110 g. We need to raise the temperature of this water from 0°C to 5°C. The specific heat of water is 1 cal/g°C. The heat required to raise the temperature (Q2) is calculated as:

\[ Q_2 = \text{mass of water} \times \text{specific heat} \times \Delta T = 110 \, \text{g} \times 1 \, \text{cal/g°C} \times 5 \, \text{°C} = 550 \, \text{cal} \]

### Step 3: Total heat required
Now, we add the heat required to melt the ice and the heat required to raise the temperature of the water:

\[ Q_{\text{total}} = Q_1 + Q_2 = 800 \, \text{cal} + 550 \, \text{cal} = 1350 \, \text{cal} \]

### Step 4: Calculate the heat released by the steam
When steam condenses into water, it releases heat. The latent heat of vaporization of steam is 540 cal/g. The steam will also cool from 100°C to 5°C. The heat released by the steam (Q_steam) can be calculated as:

\[ Q_{\text{steam}} = \text{mass of steam} \times \text{latent heat of vaporization} + \text{mass of steam} \times \text{specific heat} \times \Delta T \]

Let the mass of steam be \( m \). The heat released by the steam can be expressed as:

\[ Q_{\text{steam}} = m \times 540 \, \text{cal/g} + m \times 1 \, \text{cal/g°C} \times (100 - 5) \, \text{°C} \]
\[ Q_{\text{steam}} = m \times 540 + m \times 95 = m \times (540 + 95) = m \times 635 \, \text{cal} \]

### Step 5: Set the heat released equal to the heat required
Since the heat released by the steam must equal the total heat required to melt the ice and raise the temperature of the water, we set up the equation:

\[ m \times 635 = 1350 \]

### Step 6: Solve for the mass of steam
Now, we can solve for \( m \):

\[ m = \frac{1350}{635} \approx 2.12 \, \text{g} \]

Thus, the mass of steam required is approximately **2.12 g**.