`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, we need to determine the final amount of water in the mixture when 2 kg of ice at -20°C is mixed with 5 kg of water at 20°C. We will calculate the heat exchange between the ice and the water until thermal equilibrium is reached.

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

The formula for heat (Q) is given by:
\[ Q = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of ice = 2 kg = 2000 g (since 1 kg = 1000 g)
- \( c \) = specific heat of ice = 0.5 cal/g°C
- \( \Delta T \) = change in temperature = 0°C - (-20°C) = 20°C

Now, substituting the values:
\[ Q_{\text{ice}} = 2000 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} \]
\[ Q_{\text{ice}} = 2000 \cdot 0.5 \cdot 20 = 20000 \, \text{cal} \]

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

Using the same formula:
- \( m \) = mass of water = 5 kg = 5000 g
- \( c \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = 20°C - 0°C = 20°C

Now, substituting the values:
\[ Q_{\text{water}} = 5000 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 20 \, \text{°C} \]
\[ Q_{\text{water}} = 5000 \cdot 1 \cdot 20 = 100000 \, \text{cal} \]

### Step 3: Determine if the heat from the water is sufficient to raise the ice to 0°C and melt it.

The total heat required to raise the ice to 0°C and then melt it is:
1. Heat to raise the temperature of ice to 0°C = 20000 cal (calculated in Step 1)
2. Heat to melt the ice at 0°C:
\[ Q_{\text{melt}} = m \cdot L_f \]
Where:
- \( L_f \) = latent heat of fusion of ice = 80 cal/g
- \( m \) = mass of ice = 2000 g

Thus,
\[ Q_{\text{melt}} = 2000 \, \text{g} \cdot 80 \, \text{cal/g} = 160000 \, \text{cal} \]

Total heat required for ice:
\[ Q_{\text{total}} = Q_{\text{ice}} + Q_{\text{melt}} = 20000 \, \text{cal} + 160000 \, \text{cal} = 180000 \, \text{cal} \]

### Step 4: Compare the heat available from the water and the heat required for the ice.

The heat released by the water is 100000 cal, which is less than the heat required (180000 cal) to raise the temperature of the ice to 0°C and melt it. Therefore, not all the ice will melt.

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

The heat available from the water is 100000 cal. First, we will use this to raise the ice to 0°C:
- Heat used to raise ice to 0°C = 20000 cal (already calculated).
- Remaining heat available for melting:
\[ Q_{\text{remaining}} = 100000 \, \text{cal} - 20000 \, \text{cal} = 80000 \, \text{cal} \]

Now, calculate how much ice can melt with the remaining heat:
\[ m_{\text{melted}} = \frac{Q_{\text{remaining}}}{L_f} = \frac{80000 \, \text{cal}}{80 \, \text{cal/g}} = 1000 \, \text{g} = 1 \, \text{kg} \]

### Step 6: Calculate the final amount of water in the mixture.

Initially, we had:
- Water = 5 kg
- Ice = 2 kg

After the process:
- Melted ice = 1 kg (which becomes water)
- Remaining ice = 2 kg - 1 kg = 1 kg (which remains as ice)

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 will be **6 kg**.

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