200 g of ice at `–20°C` is mixed with 500 g of water `20°C` in an insulating vessel. Final mass of water in vessel is (specific heat of ice `– 0.5 cal g^(–1°)C^(–1)`)

Options

A. 700 g
B. 600 g
C. 400 g
D. 200 g

Answer

To solve the problem of mixing 200 g of ice at -20°C with 500 g of water at 20°C, we need to calculate the final mass of water in the vessel after the thermal equilibrium is reached. Here’s a step-by-step solution:

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

The formula for the heat required (Q) to change the temperature is:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m \) = mass of ice = 200 g
- \( s \) = specific heat of ice = 0.5 cal/g°C
- \( \Delta T \) = change in temperature = 0 - (-20) = 20°C

Calculating:
\[ Q = 200 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 2000 \, \text{cal} = 2 \, \text{kcal} \]

### Step 2: Calculate the heat required to melt the ice at 0°C.

The latent heat of fusion for ice is 80 cal/g. The heat required to melt the ice is given by:
\[ Q = m \cdot L_f \]
where:
- \( L_f \) = latent heat of fusion = 80 cal/g

Calculating:
\[ Q = 200 \, \text{g} \cdot 80 \, \text{cal/g} = 16000 \, \text{cal} = 16 \, \text{kcal} \]

### Step 3: Total heat required to convert ice at -20°C to water at 0°C.

Total heat required:
\[ Q_{total} = Q_{heating} + Q_{melting} = 2 \, \text{kcal} + 16 \, \text{kcal} = 18 \, \text{kcal} \]

### Step 4: Calculate the heat lost by the water as it cools from 20°C to 0°C.

The heat lost by the water can be calculated using the same formula:
\[ Q = m \cdot s \cdot \Delta T \]
where:
- \( m \) = mass of water = 500 g
- \( s \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = 20°C - 0°C = 20°C

Calculating:
\[ Q = 500 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 10000 \, \text{cal} = 10 \, \text{kcal} \]

### Step 5: Determine if the heat lost by water is sufficient to melt the ice.

The heat lost by the water (10 kcal) is less than the total heat required to convert the ice to water (18 kcal). Therefore, not all the ice will melt.

### Step 6: Calculate the remaining heat after melting some of the ice.

The heat available from the water is 10 kcal. The heat required to raise the ice to 0°C is 2 kcal, leaving:
\[ Q_{remaining} = 10 \, \text{kcal} - 2 \, \text{kcal} = 8 \, \text{kcal} \]

### Step 7: Calculate how much ice can be melted with the remaining heat.

Using the latent heat of fusion:
\[ Q_{remaining} = m \cdot L_f \]
\[ 8 \, \text{kcal} = m \cdot 80 \, \text{cal/g} \]
Converting kcal to cal:
\[ 8 \, \text{kcal} = 8000 \, \text{cal} \]
Thus:
\[ 8000 \, \text{cal} = m \cdot 80 \, \text{cal/g} \]
\[ m = \frac{8000 \, \text{cal}}{80 \, \text{cal/g}} = 100 \, \text{g} \]

### Step 8: Calculate the final mass of water in the vessel.

Initially, there were 500 g of water and 200 g of ice. After melting 100 g of ice, the total mass of water is:
\[ \text{Final mass of water} = 500 \, \text{g} + (200 \, \text{g} - 100 \, \text{g}) = 600 \, \text{g} \]

### Final Answer:
The final mass of water in the vessel is **600 g**.

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