A metal piece weighing `15 g` is heated to `100^(@)C` and then immersed in a mixture of ice and water at the thermal equilibrium. The volume of the mixture is found to be reduced by `0.15 cm^(3)` with the temperature of mixture remaining constant. Find the specific heat of the metal. Given specific gravity of ice `= 0.92`, specific gravity of water at `0^(@)C = 1.0`, latent heat of fusion of ice `= 80 cal-g^(-1)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a metal piece weighing 15 g that is heated to 100°C and then immersed in a mixture of ice and water. The volume of the mixture is reduced by 0.15 cm³, and we need to find the specific heat of the metal.

### Step 2: Calculate the mass of ice melted
The reduction in volume of the mixture (0.15 cm³) indicates that some ice has melted. We can use the specific gravity of ice to find the mass of the ice that melted.

- Specific gravity of ice = 0.92
- Density of ice = 0.92 g/cm³

Using the formula for mass:
\[ \text{Mass of ice melted} = \text{Volume} \times \text{Density} = 0.15 \, \text{cm}^3 \times 0.92 \, \text{g/cm}^3 = 0.138 \, \text{g} \]

### Step 3: Calculate the heat lost by the ice
The heat lost by the ice as it melts can be calculated using the latent heat of fusion of ice.

- Latent heat of fusion of ice = 80 cal/g

Using the formula:
\[ Q_{\text{lost}} = m \times L = 0.138 \, \text{g} \times 80 \, \text{cal/g} = 11.04 \, \text{cal} \]

### Step 4: Calculate the heat gained by the metal
The heat gained by the metal can be expressed as:
\[ Q_{\text{gained}} = m_{\text{metal}} \times s \times \Delta T \]
Where:
- \( m_{\text{metal}} = 15 \, \text{g} \)
- \( \Delta T = 100 - 0 = 100 \, \text{°C} \) (since the final temperature of the mixture is 0°C)

Thus,
\[ Q_{\text{gained}} = 15 \, \text{g} \times s \times 100 \, \text{°C} \]

### Step 5: Set heat gained equal to heat lost
From the principle of calorimetry, we know that the heat gained by the metal is equal to the heat lost by the ice:
\[ Q_{\text{gained}} = Q_{\text{lost}} \]
Substituting the values we have:
\[ 15 \, \text{g} \times s \times 100 \, \text{°C} = 11.04 \, \text{cal} \]

### Step 6: Solve for specific heat \( s \)
Rearranging the equation to solve for \( s \):
\[ s = \frac{11.04 \, \text{cal}}{15 \, \text{g} \times 100 \, \text{°C}} = \frac{11.04}{1500} \approx 0.00736 \, \text{cal/g°C} \]

### Step 7: Convert to appropriate units
To express the specific heat in a more standard form:
\[ s \approx 0.00736 \, \text{cal/g°C} \approx 0.0736 \, \text{cal/g°C} \]

### Final Answer
The specific heat of the metal is approximately \( 0.0736 \, \text{cal/g°C} \).

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