A metal ball of specific gravity `4.5` and specific heat `0.1 cal//gm-"^(@)C` is placed on a large slab of ice at `0^(@)C`. Half of the ball sinks in the ice. The initial temperature of the ball is:- (Latent heat capacity of ice `=80 cal//g`, specific gravity of ice `=0.9)`

Options

A. `100"^(@)C`
B. `90"^(@)C`
C. `80"^(@)C`
D. `70"^(@)C`

Explanation

To solve the problem, we will use the principle of calorimetry, which states that the heat lost by the metal ball will be equal to the heat gained by the ice. Here's a step-by-step solution:

### Step 1: Define the variables
- Let the mass of the metal ball be \( m \).
- The specific gravity of the metal ball is \( 4.5 \), which means its density \( \rho_{ball} = 4.5 \, \text{g/cm}^3 \).
- The specific heat of the metal ball is \( s = 0.1 \, \text{cal/g}^\circ C \).
- The latent heat of fusion of ice is \( L = 80 \, \text{cal/g} \).
- The specific gravity of ice is \( 0.9 \), which means its density \( \rho_{ice} = 0.9 \, \text{g/cm}^3 \).

### Step 2: Calculate the mass of the ball
The volume \( V \) of the ball can be expressed in terms of its mass and density:
\[ m = V \cdot \rho_{ball} \]

### Step 3: Heat lost by the metal ball
The heat lost by the metal ball when it cools down from its initial temperature \( T \) to \( 0^\circ C \) is given by:
\[ Q_{lost} = m \cdot s \cdot (T - 0) = m \cdot s \cdot T \]

### Step 4: Heat gained by the ice
Since half of the ball sinks into the ice, the mass of the ice that melts can be calculated as:
\[ m_{ice} = \frac{V}{2} \cdot \rho_{ice} \]
The heat gained by the ice as it melts is given by:
\[ Q_{gained} = m_{ice} \cdot L = \left(\frac{V}{2} \cdot \rho_{ice}\right) \cdot L \]

### Step 5: Set the heat lost equal to the heat gained
According to the principle of conservation of energy:
\[ Q_{lost} = Q_{gained} \]
Substituting the expressions for heat:
\[ m \cdot s \cdot T = \left(\frac{V}{2} \cdot \rho_{ice}\right) \cdot L \]

### Step 6: Substitute mass in terms of volume and density
From Step 2, we know:
\[ m = V \cdot \rho_{ball} \]
Substituting this into the equation:
\[ (V \cdot \rho_{ball}) \cdot s \cdot T = \left(\frac{V}{2} \cdot \rho_{ice}\right) \cdot L \]

### Step 7: Cancel out the volume \( V \)
Assuming \( V \neq 0 \):
\[ \rho_{ball} \cdot s \cdot T = \frac{1}{2} \cdot \rho_{ice} \cdot L \]

### Step 8: Substitute known values
Substituting the known values:
- \( \rho_{ball} = 4.5 \, \text{g/cm}^3 \)
- \( s = 0.1 \, \text{cal/g}^\circ C \)
- \( \rho_{ice} = 0.9 \, \text{g/cm}^3 \)
- \( L = 80 \, \text{cal/g} \)

The equation becomes:
\[ 4.5 \cdot 0.1 \cdot T = \frac{1}{2} \cdot 0.9 \cdot 80 \]

### Step 9: Simplify and solve for \( T \)
Calculating the right side:
\[ 4.5 \cdot 0.1 \cdot T = 0.45 \cdot 80 \]
\[ 4.5 \cdot 0.1 \cdot T = 36 \]
\[ T = \frac{36}{0.45} = 80^\circ C \]

### Conclusion
The initial temperature of the ball is \( 80^\circ C \).

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