`2kg` ice at `-20"^(@)C` is mixed with `5kg` water at `20"^(@)C`. Then final amount of water in the mixture will be: [specific heat of ice `=0.5 cal//gm "^(@)C`, Specific heat of water `=1 cal//gm"^(@)C`, Latent heat of fusion of ice `= 80 cal//gm]`

Options

A. `6kg`
B. `7kg`
C. `3.5 kg`
D. `5kg`

Answer

To solve the problem of mixing 2 kg of ice at -20°C with 5 kg of water at 20°C, we need to follow these steps:

### Step 1: Calculate the heat absorbed by the ice to reach 0°C
The heat absorbed by the ice to raise its temperature from -20°C to 0°C can be calculated using the formula:
\[ Q_1 = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of ice = 2 kg = 2000 g (since 1 kg = 1000 g)
- \( s \) = specific heat of ice = 0.5 cal/g°C
- \( \Delta T \) = change in temperature = 0 - (-20) = 20°C

Calculating \( Q_1 \):
\[ Q_1 = 2000 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 20000 \, \text{cal} \]

### Step 2: Calculate the heat required to convert ice at 0°C to water at 0°C
The heat required to convert the ice at 0°C to water at 0°C is given by:
\[ Q_2 = m \cdot L \]
Where:
- \( L \) = latent heat of fusion of ice = 80 cal/g

Calculating \( Q_2 \):
\[ Q_2 = 2000 \, \text{g} \cdot 80 \, \text{cal/g} = 160000 \, \text{cal} \]

### Step 3: Calculate the total heat absorbed by the ice
The total heat absorbed by the ice is:
\[ Q_{ice} = Q_1 + Q_2 = 20000 \, \text{cal} + 160000 \, \text{cal} = 180000 \, \text{cal} \]

### Step 4: Calculate the heat released by the water as it cools to 0°C
The heat released by the water as it cools from 20°C to 0°C is given by:
\[ Q_{water} = m \cdot s \cdot \Delta T \]
Where:
- \( m \) = mass of water = 5 kg = 5000 g
- \( s \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = 20°C - 0°C = 20°C

Calculating \( Q_{water} \):
\[ Q_{water} = 5000 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 100000 \, \text{cal} \]

### Step 5: Determine the heat balance
Now we compare the heat absorbed by the ice and the heat released by the water:
- Heat absorbed by the ice: \( 180000 \, \text{cal} \)
- Heat released by the water: \( 100000 \, \text{cal} \)

Since the heat absorbed by the ice (180000 cal) is greater than the heat released by the water (100000 cal), not all the ice will melt.

### Step 6: Calculate how much ice melts
The heat released by the water is used to first raise the temperature of the ice to 0°C and then to melt some of it. The heat used to raise the temperature of the ice is 20000 cal, leaving:
\[ 100000 \, \text{cal} - 20000 \, \text{cal} = 80000 \, \text{cal} \]
This remaining heat is used to melt the ice:
\[ Q_{melt} = x \cdot L \]
Where \( x \) is the mass of ice melted. Rearranging gives:
\[ x = \frac{Q_{melt}}{L} = \frac{80000 \, \text{cal}}{80 \, \text{cal/g}} = 1000 \, \text{g} = 1 \, \text{kg} \]

### Step 7: Calculate the final amounts of water and ice
- Initial water: 5 kg
- Melted ice: 1 kg
- Remaining ice: 2 kg - 1 kg = 1 kg

Thus, the final amount of water in the mixture is:
\[ \text{Final water} = 5 \, \text{kg} + 1 \, \text{kg} = 6 \, \text{kg} \]

### Final Answer
The final amount of water in the mixture is **6 kg**.

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