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 0.5`
`kcal//kg//^@C` while the latent heat of fusion of ice is `80kcal//kg`

Options

A. `7kg`
B. `5kg`
C. `4kg`
D. `2kg`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the heat lost by the water
The first step is to calculate the heat lost by the 5 kg of water when it cools down from 20°C to 0°C.

Using the formula:
\[ Q = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of water = 5 kg
- \( c \) = specific heat of water = 1 kcal/kg°C
- \( \Delta T \) = change in temperature = \( 20°C - 0°C = 20°C \)

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

### Step 2: Calculate the heat gained by the ice to reach 0°C
Next, we need to calculate the heat gained by the 2 kg of ice to raise its temperature from -20°C to 0°C.

Using the same formula:
\[ Q = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of ice = 2 kg
- \( c \) = specific heat of ice = 0.5 kcal/kg°C
- \( \Delta T \) = change in temperature = \( 0°C - (-20°C) = 20°C \)

Calculating:
\[ Q_{ice} = 2 \, \text{kg} \cdot 0.5 \, \text{kcal/kg°C} \cdot 20 \, \text{°C} = 20 \, \text{kcal} \]

### Step 3: Determine the heat available for melting the ice
Now, we find the remaining heat after the ice has reached 0°C. The heat lost by the water is 100 kcal, and the heat gained by the ice to reach 0°C is 20 kcal.

Remaining heat:
\[ Q_{remaining} = Q_{water} - Q_{ice} = 100 \, \text{kcal} - 20 \, \text{kcal} = 80 \, \text{kcal} \]

### Step 4: Calculate how much ice can melt with the remaining heat
The remaining heat will be used to melt the ice. The latent heat of fusion of ice is given as 80 kcal/kg.

Using the formula:
\[ Q = m \cdot L_f \]

Where:
- \( L_f \) = latent heat of fusion = 80 kcal/kg
- \( Q \) = remaining heat = 80 kcal

Let \( x \) be the mass of ice that melts:
\[ 80 \, \text{kcal} = x \cdot 80 \, \text{kcal/kg} \]

Solving for \( x \):
\[ x = \frac{80 \, \text{kcal}}{80 \, \text{kcal/kg}} = 1 \, \text{kg} \]

### Step 5: Calculate the final mass of water
Initially, we had 5 kg of water and 2 kg of ice. After melting, 1 kg of ice becomes water.

Final mass of water:
\[ \text{Total water} = 5 \, \text{kg (from water)} + 1 \, \text{kg (from melted ice)} = 6 \, \text{kg} \]

### Conclusion
The final mass of water remaining in the container is **6 kg**.

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