`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:

Options

A. 20 g of water
B. 20 g of ice
C. 10 g ice and 10 g of water
D. 5 g ice and 15 g of water

Explanation

To solve the problem step by step, we need to analyze the heat transfer between the ice and the water until thermal equilibrium is reached.

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

The specific heat of ice is given as \( \frac{1}{2} \) cal/g°C.

- Mass of ice, \( m_{ice} = 10 \) g
- Temperature change, \( \Delta T = 0 - (-20) = 20 \)°C

Using the formula for heat transfer:
\[ Q_1 = m_{ice} \cdot c_{ice} \cdot \Delta T \]
\[ Q_1 = 10 \, \text{g} \cdot \frac{1}{2} \, \text{cal/g°C} \cdot 20 \, \text{°C} = 100 \, \text{cal} \]

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

The latent heat of fusion of ice is \( L = 80 \) cal/g.

Using the formula for phase change:
\[ Q_2 = m_{ice} \cdot L \]
\[ Q_2 = 10 \, \text{g} \cdot 80 \, \text{cal/g} = 800 \, \text{cal} \]

### Step 3: Calculate the total heat absorbed by the ice.

\[ Q_{ice} = Q_1 + Q_2 = 100 \, \text{cal} + 800 \, \text{cal} = 900 \, \text{cal} \]

### Step 4: Calculate the heat released by the water as it cools from 10°C to 0°C.

The specific heat of water is \( 1 \) cal/g°C.

- Mass of water, \( m_{water} = 10 \) g
- Temperature change, \( \Delta T = 10 - 0 = 10 \)°C

Using the formula for heat transfer:
\[ Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T \]
\[ Q_{water} = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 10 \, \text{°C} = 100 \, \text{cal} \]

### Step 5: Determine the net heat exchange.

The ice requires 900 cal to convert to water, while the water can only release 100 cal.

- Heat available from water: \( 100 \, \text{cal} \)
- Heat required by ice: \( 900 \, \text{cal} \)

The ice will absorb all the heat from the water, but it will not be enough to completely convert all the ice into water.

### Step 6: Calculate the extra heat absorbed by the ice.

The extra heat absorbed by the ice after the water has cooled down is:
\[ Q_{extra} = Q_{ice} - Q_{water} = 900 \, \text{cal} - 100 \, \text{cal} = 800 \, \text{cal} \]

### Step 7: Determine how much ice can be melted with the extra heat.

Using the latent heat of fusion:
\[ Q_{extra} = m_{ice} \cdot L \]
\[ 800 \, \text{cal} = m_{ice} \cdot 80 \, \text{cal/g} \]
\[ m_{ice} = \frac{800 \, \text{cal}}{80 \, \text{cal/g}} = 10 \, \text{g} \]

### Step 8: Conclusion

Since all the ice (10 g) can be melted, and the water (10 g) will remain at 0°C, the final equilibrium state will contain:

- 10 g of water (from the melted ice)
- 10 g of water (original water)

Thus, the calorimeter will contain **10 g of ice and 10 g of water** at equilibrium.