A 50gram lead bullet, specific heat 0.02 is initially at `30^(@)` C. It is fired vertically upwards with a speed of `840 m// s` . On returning to the starting level, it strikes a cake of ice at `0^(@)C`. How much ice is melted ? Assume that all energy is spent in melting ice only. Take latent heat of ice `= 80 cal. //` gram.

Explanation

To solve the problem step by step, we will follow the process outlined in the video transcript while providing detailed explanations for each step.

### Step 1: Identify the Given Data
We start by noting the given data from the problem:
- Mass of the bullet, \( m = 50 \, \text{grams} = 0.05 \, \text{kg} \)
- Specific heat of lead, \( c = 0.02 \, \text{cal/g°C} \)
- Initial temperature of the bullet, \( T_b = 30 \, \text{°C} \)
- Speed of the bullet, \( v = 840 \, \text{m/s} \)
- Temperature of the ice, \( T_i = 0 \, \text{°C} \)
- Latent heat of ice, \( L = 80 \, \text{cal/g} \)

### Step 2: Calculate the Kinetic Energy of the Bullet
The kinetic energy (KE) of the bullet when it strikes the ice can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 0.05 \, \text{kg} \times (840 \, \text{m/s})^2 \]
Calculating this gives:
\[ KE = \frac{1}{2} \times 0.05 \times 705600 = 17640 \, \text{J} \]

### Step 3: Convert Kinetic Energy to Calories
To convert joules to calories, we use the conversion factor \( 1 \, \text{J} = 0.239 \, \text{cal} \):
\[ KE = 17640 \, \text{J} \times \frac{1 \, \text{cal}}{4.184 \, \text{J}} \approx 4200 \, \text{cal} \]

### Step 4: Calculate Heat Lost by the Bullet to Cool Down
The bullet will lose heat as it cools down from \( 30 \, \text{°C} \) to \( 0 \, \text{°C} \). The heat lost (Q1) can be calculated using:
\[ Q_1 = mc\Delta T \]
Where \( \Delta T = T_b - T_i = 30 - 0 = 30 \, \text{°C} \):
\[ Q_1 = 50 \, \text{g} \times 0.02 \, \text{cal/g°C} \times 30 \, \text{°C} = 30 \, \text{cal} \]

### Step 5: Calculate Remaining Heat for Melting Ice
The remaining heat (Q_left) that is available to melt the ice is:
\[ Q_{\text{left}} = KE - Q_1 = 4200 \, \text{cal} - 30 \, \text{cal} = 4170 \, \text{cal} \]

### Step 6: Calculate the Mass of Ice Melted
The heat required to melt ice is given by:
\[ Q_{\text{left}} = m_{\text{ice}} \times L \]
Rearranging for \( m_{\text{ice}} \):
\[ m_{\text{ice}} = \frac{Q_{\text{left}}}{L} = \frac{4170 \, \text{cal}}{80 \, \text{cal/g}} \approx 52.125 \, \text{g} \]

### Final Answer
The mass of ice melted is approximately \( 52.125 \, \text{grams} \).

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