`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 gm` of water
B. `20 gm` of ice
C. `10 gm` ice and `10 gm` of water
D. `5 gm` ice and `15 gm` of water

Explanation

To solve the problem, we need to analyze the heat exchange between the ice and the water until thermal equilibrium is reached. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Mass of ice, \( m_{\text{ice}} = 10 \, \text{g} \)
- Initial temperature of ice, \( T_{\text{ice initial}} = -20^\circ C \)
- Mass of water, \( m_{\text{water}} = 10 \, \text{g} \)
- Initial temperature of water, \( T_{\text{water initial}} = 10^\circ C \)
- Specific heat of water, \( c_{\text{water}} = 1 \, \text{cal/g}^\circ C \)
- Specific heat of ice, \( c_{\text{ice}} = \frac{1}{2} \, \text{cal/g}^\circ C \) (since it is given that the specific heat of water is twice that of ice)

### Step 2: Calculate the heat lost by the water as it cools down to \( 0^\circ C \)
The heat lost by the water can be calculated using the formula:
\[ Q_1 = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{water initial}} - T_{\text{final}}) \]
Substituting the known values:
\[ Q_1 = 10 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot (10^\circ C - 0^\circ C) = 10 \cdot 1 \cdot 10 = 100 \, \text{cal} \]

### Step 3: Calculate the heat required to warm the ice from \( -20^\circ C \) to \( 0^\circ C \)
The heat required to raise the temperature of the ice can be calculated using:
\[ Q_2 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot (T_{\text{final}} - T_{\text{ice initial}}) \]
Substituting the known values:
\[ Q_2 = 10 \, \text{g} \cdot \frac{1}{2} \, \text{cal/g}^\circ C \cdot (0^\circ C - (-20^\circ C)) = 10 \cdot \frac{1}{2} \cdot 20 = 100 \, \text{cal} \]

### Step 4: Compare the heat lost and gained
- Heat lost by water, \( Q_1 = 100 \, \text{cal} \)
- Heat gained by ice, \( Q_2 = 100 \, \text{cal} \)

Since \( Q_1 = Q_2 \), the heat lost by the water is equal to the heat gained by the ice.

### Step 5: Conclusion
At thermal equilibrium, all the ice will have reached \( 0^\circ C \) and will not have melted, and the water will also be at \( 0^\circ C \). Therefore, the calorimeter will contain:
- 10 grams of ice at \( 0^\circ C \)
- 10 grams of water at \( 0^\circ C \)

### Final Answer
The calorimeter will contain **10 grams of ice and 10 grams of water, both at \( 0^\circ C \)**.

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