An iron ball of mass 0.2 kg is heated to `10^(@)C` and put into a block of ice at `0^(@)C`. 25 g of ice melts. If the latent heat of fusion of ice is `80 cal g^(-1)`, then the specific heat of iron in `cal g^(-1^(@))C` is

Options

A. 1
B. 0.1
C. 0.8
D. 0.008

Explanation

To solve the problem, we need to apply the principle of conservation of energy, which states that the heat lost by the iron ball will equal the heat gained by the ice.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of the iron ball, \( m_{iron} = 0.2 \, \text{kg} = 200 \, \text{g} \)
- Initial temperature of the iron ball, \( T_{initial} = 10^\circ C \)
- Final temperature of the ice, \( T_{final} = 0^\circ C \)
- Mass of the melted ice, \( m_{ice} = 25 \, \text{g} \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)

2. **Calculate the heat gained by the ice:**
The heat gained by the ice when it melts can be calculated using the formula:
\[ Q_{gained} = m_{ice} \times L \]
Substituting the values:
\[ Q_{gained} = 25 \, \text{g} \times 80 \, \text{cal/g} = 2000 \, \text{cal} \]

3. **Set up the equation for heat lost by the iron ball:**
The heat lost by the iron ball can be calculated using the formula:
\[ Q_{lost} = m_{iron} \times s \times \Delta T \]
where \( s \) is the specific heat of iron and \( \Delta T \) is the change in temperature. Here, \( \Delta T = T_{initial} - T_{final} = 10^\circ C - 0^\circ C = 10^\circ C \).
Therefore, we have:
\[ Q_{lost} = 200 \, \text{g} \times s \times 10 \]

4. **Apply the conservation of energy principle:**
According to the conservation of energy:
\[ Q_{lost} = Q_{gained} \]
Substituting the values we have:
\[ 200 \, \text{g} \times s \times 10 = 2000 \, \text{cal} \]

5. **Solve for the specific heat \( s \):**
Rearranging the equation to find \( s \):
\[ s = \frac{2000 \, \text{cal}}{200 \, \text{g} \times 10} \]
Simplifying:
\[ s = \frac{2000}{2000} = 1 \, \text{cal/g}^\circ C \]

### Final Answer:
The specific heat of iron is \( 1 \, \text{cal/g}^\circ C \).

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