2kg of ice at `-20^@C` is mixed with 5kg of water at `20^@C` in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water & ice are `1kcal//kg//^@C` and `0.5`
`kcal//kg//^@C` while the latent heat of fusion of ice is `80kcal//kg`

Options

A. 7 kg
B. 6 kg
C. 4 kg
D. 2 kg

Explanation

To solve the problem of mixing 2 kg of ice at -20°C with 5 kg of water at 20°C in an insulating vessel, we will follow these steps:

### Step 1: Calculate the heat required to raise the temperature of ice from -20°C to 0°C.

The specific heat of ice is given as \(0.5 \, \text{kcal/kg°C}\).

\[ Q_1 = m \cdot c \cdot \Delta T \]
Where:
- \(m = 2 \, \text{kg}\) (mass of ice)
- \(c = 0.5 \, \text{kcal/kg°C}\) (specific heat of ice)
- \(\Delta T = 0 - (-20) = 20°C\) (change in temperature)

Calculating \(Q_1\):
\[ Q_1 = 2 \, \text{kg} \cdot 0.5 \, \text{kcal/kg°C} \cdot 20°C = 20 \, \text{kcal} \]

### Step 2: Calculate the heat released by the water as it cools from 20°C to 0°C.

The specific heat of water is given as \(1 \, \text{kcal/kg°C}\).

\[ Q_2 = m \cdot c \cdot \Delta T \]
Where:
- \(m = 5 \, \text{kg}\) (mass of water)
- \(c = 1 \, \text{kcal/kg°C}\) (specific heat of water)
- \(\Delta T = 20 - 0 = 20°C\) (change in temperature)

Calculating \(Q_2\):
\[ Q_2 = 5 \, \text{kg} \cdot 1 \, \text{kcal/kg°C} \cdot 20°C = 100 \, \text{kcal} \]

### Step 3: Calculate the heat required to convert the ice at 0°C to water at 0°C.

The latent heat of fusion of ice is given as \(80 \, \text{kcal/kg}\).

\[ Q_3 = m \cdot L_f \]
Where:
- \(m = 2 \, \text{kg}\) (mass of ice)
- \(L_f = 80 \, \text{kcal/kg}\) (latent heat of fusion)

Calculating \(Q_3\):
\[ Q_3 = 2 \, \text{kg} \cdot 80 \, \text{kcal/kg} = 160 \, \text{kcal} \]

### Step 4: Determine the heat balance.

The heat gained by the ice to reach 0°C and then to convert to water must equal the heat lost by the water cooling down to 0°C.

1. Heat gained by ice to reach 0°C: \(Q_1 = 20 \, \text{kcal}\)
2. Heat lost by water cooling to 0°C: \(Q_2 = 100 \, \text{kcal}\)

Since the total heat required to convert all the ice to water is \(Q_1 + Q_3 = 20 \, \text{kcal} + 160 \, \text{kcal} = 180 \, \text{kcal}\), and the heat available from the water is only \(100 \, \text{kcal}\), not all the ice can melt.

### Step 5: Calculate how much ice can melt with the available heat.

The heat available after raising the ice to 0°C is:
\[ Q_{\text{available}} = Q_2 - Q_1 = 100 \, \text{kcal} - 20 \, \text{kcal} = 80 \, \text{kcal} \]

This heat will be used to melt some of the ice:
\[ Q_{\text{melt}} = m \cdot L_f \implies 80 \, \text{kcal} = m \cdot 80 \, \text{kcal/kg} \]
Solving for \(m\):
\[ m = \frac{80 \, \text{kcal}}{80 \, \text{kcal/kg}} = 1 \, \text{kg} \]

### Step 6: Calculate the final mass of water.

The final mass of water in the container is the sum of the original mass of water and the mass of ice that melted:
\[ \text{Final mass of water} = 5 \, \text{kg} + 1 \, \text{kg} = 6 \, \text{kg} \]

### Final Answer:
The final mass of water remaining in the container is **6 kg**.

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