Two identical pieces of ice fly towards each other with equal and opposite velocities and are converted into water upon impact. If their initial temperature is `-12^(@)C`, then the minimum possible velocity of either piece before impact is `(S_("ice") = 0.50 "cal" gm^(-1) ""^(@)C^(-1), 1"cal" = 4.2J)`

Options

A. (a)850 m/sec
B. (b)600m/sec
C. (c)1000m/sec
D. (d)500m/sec

Explanation

To solve the problem, we need to calculate the minimum possible velocity of either piece of ice before they collide and convert into water. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Energy Conversion
When the two pieces of ice collide, their kinetic energy is converted into thermal energy, which will first raise the temperature of the ice to 0°C and then convert the ice at 0°C to water at 0°C.

### Step 2: Write the Energy Balance Equation
The total kinetic energy lost by the two pieces of ice will equal the energy required to:
1. Raise the temperature of the ice from -12°C to 0°C.
2. Melt the ice at 0°C into water at 0°C.

The equation can be written as:
\[ \text{Loss in Kinetic Energy} = \text{Energy to raise temperature} + \text{Energy to melt ice} \]

### Step 3: Calculate the Energy Required
1. **Energy to raise temperature**:
\[ Q_1 = m \cdot s \cdot \Delta T \]
where:
- \( m \) = mass of one piece of ice (we can assume \( m = 1 \) g for simplicity)
- \( s \) = specific heat of ice = 0.50 cal/g°C
- \( \Delta T = 12°C \) (from -12°C to 0°C)

Thus,
\[ Q_1 = 1 \cdot 0.50 \cdot 12 = 6 \text{ cal} \]

2. **Energy to melt ice**:
\[ Q_2 = m \cdot L \]
where \( L \) is the latent heat of fusion of ice, which is approximately 80 cal/g.

Thus,
\[ Q_2 = 1 \cdot 80 = 80 \text{ cal} \]

### Step 4: Total Energy Required
The total energy required to convert the ice to water is:
\[ Q_{\text{total}} = Q_1 + Q_2 = 6 + 80 = 86 \text{ cal} \]

### Step 5: Kinetic Energy Calculation
The kinetic energy of each piece of ice before collision is given by:
\[ KE = \frac{1}{2} m v^2 \]
For two pieces of ice, the total kinetic energy is:
\[ KE_{\text{total}} = 2 \cdot \frac{1}{2} m v^2 = m v^2 \]

### Step 6: Set Kinetic Energy Equal to Energy Required
Setting the kinetic energy equal to the total energy required:
\[ m v^2 = 86 \text{ cal} \]

### Step 7: Convert Calories to Joules
Since \( 1 \text{ cal} = 4.2 \text{ J} \):
\[ 86 \text{ cal} = 86 \times 4.2 \text{ J} = 361.2 \text{ J} \]

### Step 8: Solve for Velocity
Assuming \( m = 1 \text{ g} = 0.001 \text{ kg} \):
\[ 0.001 v^2 = 361.2 \implies v^2 = \frac{361.2}{0.001} = 361200 \implies v = \sqrt{361200} \approx 600 \text{ m/s} \]

### Final Answer
The minimum possible velocity of either piece of ice before impact is approximately **600 m/s**.

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