A vessel contains `10^(–1) kg` of ice at `0^(@)C`. Now steam is passed into the vessel to melt ice. Neglecting the thermal capacity of the vessel, find the mass of water in the vessel when all ice melts into water.

Options

A. 12.5 gm
B. 100 gm
C. 112.5 gm
D. 125 gm

Answer

To solve the problem, we need to find the mass of water in the vessel when all the ice melts into water after steam is passed into it. We'll break down the solution step by step.

### Step 1: Determine the mass of ice
The mass of ice in the vessel is given as:
\[ m_{\text{ice}} = 10^{-1} \text{ kg} = 0.1 \text{ kg} = 100 \text{ g} \]

### Step 2: Calculate the heat required to melt the ice
The heat required to melt the ice (Q_melt) can be calculated using the formula:
\[ Q_{\text{melt}} = m_{\text{ice}} \times L_f \]
where \(L_f\) is the latent heat of fusion of ice, which is approximately \(80 \text{ cal/g}\).

Substituting the values:
\[ Q_{\text{melt}} = 100 \text{ g} \times 80 \text{ cal/g} = 8000 \text{ cal} \]

### Step 3: Determine the heat released by the steam
Let the mass of steam that condenses into water be \(m\) grams. The steam first condenses to water at \(100^\circ C\) and then cools to \(0^\circ C\).

1. **Heat released during condensation**:
The heat released when steam condenses to water is given by:
\[ Q_{\text{condense}} = m \times L_v \]
where \(L_v\) is the latent heat of vaporization of water, approximately \(540 \text{ cal/g}\).

2. **Heat released during cooling from \(100^\circ C\) to \(0^\circ C\)**:
The heat released when water cools from \(100^\circ C\) to \(0^\circ C\) is given by:
\[ Q_{\text{cool}} = m \times c \times \Delta T \]
where \(c\) is the specific heat capacity of water (\(1 \text{ cal/g}^\circ C\)) and \(\Delta T = 100^\circ C\).

So,
\[ Q_{\text{cool}} = m \times 1 \text{ cal/g}^\circ C \times 100^\circ C = 100m \text{ cal} \]

### Step 4: Total heat released by the steam
The total heat released by the steam is:
\[ Q_{\text{released}} = Q_{\text{condense}} + Q_{\text{cool}} = m \times 540 + 100m = 640m \text{ cal} \]

### Step 5: Set the heat released equal to the heat required
For all the ice to melt, the heat released by the steam must equal the heat required to melt the ice:
\[ 640m = 8000 \]

### Step 6: Solve for the mass of steam
Now, we can solve for \(m\):
\[ m = \frac{8000}{640} = 12.5 \text{ g} \]

### Step 7: Calculate the total mass of water in the vessel
The total mass of water in the vessel when all the ice has melted is the sum of the mass of melted ice and the mass of steam that has condensed into water:
\[ \text{Total mass of water} = m_{\text{ice}} + m = 100 \text{ g} + 12.5 \text{ g} = 112.5 \text{ g} \]

### Final Answer
The total mass of water in the vessel when all the ice melts is:
\[ \text{Total mass of water} = 112.5 \text{ g} \]