In an experiment to determine the specific heat of aluminium, piece of aluminimum weighing `500 g` is heated to `100 .^(@) C`. It is then quickly transferred into a copper calorimeter of mass `500 g` containing `300g` of water at `30 .^(@) C`. The final temperature of the mixture is found to be `46.8 .^(@) c` . If specific heat of copper is `0.093 cal g^-1 .^(@) C^-1`, then the specific heat aluminium is.

Options

A. `(a)0.11 cal g^-1 .^(@) C^-1`.
B. `(b)0.22 cal g^-1 .^(@) C^-1`.
C. `(c)0.33 cal g^-1 .^(@) C^-1`.
D. `(d)0.44 cal g^-1 .^(@) C^-1`.

Explanation

To determine the specific heat of aluminum in the given experiment, we can follow these steps:

### Step 1: Identify the heat gained and lost
In calorimetry, the heat gained by the cold body (copper calorimeter and water) is equal to the heat lost by the hot body (aluminum).

### Step 2: Write the equation for heat gained by the cold body
The heat gained by the cold body can be expressed as:
\[ Q_{\text{gain}} = (m_{\text{Cu}} \cdot c_{\text{Cu}} + m_{\text{water}} \cdot c_{\text{water}}) \cdot (T_f - T_i) \]
Where:
- \(m_{\text{Cu}} = 500 \, \text{g}\) (mass of copper calorimeter)
- \(c_{\text{Cu}} = 0.093 \, \text{cal/g°C}\) (specific heat of copper)
- \(m_{\text{water}} = 300 \, \text{g}\) (mass of water)
- \(c_{\text{water}} = 1 \, \text{cal/g°C}\) (specific heat of water)
- \(T_f = 46.8 \, \text{°C}\) (final temperature)
- \(T_i = 30 \, \text{°C}\) (initial temperature of water and calorimeter)

### Step 3: Calculate the change in temperature
The change in temperature for the cold body is:
\[ \Delta T = T_f - T_i = 46.8 \, \text{°C} - 30 \, \text{°C} = 16.8 \, \text{°C} \]

### Step 4: Substitute values into the heat gained equation
Now substituting the values into the equation:
\[ Q_{\text{gain}} = \left(500 \cdot 0.093 + 300 \cdot 1\right) \cdot 16.8 \]
Calculating each term:
\[ Q_{\text{gain}} = (46.5 + 300) \cdot 16.8 = 346.5 \cdot 16.8 \]
Calculating \(Q_{\text{gain}}\):
\[ Q_{\text{gain}} = 5821.2 \, \text{cal} \]

### Step 5: Write the equation for heat lost by the hot body
The heat lost by the aluminum can be expressed as:
\[ Q_{\text{lost}} = m_{\text{Al}} \cdot c_{\text{Al}} \cdot (T_i - T_f) \]
Where:
- \(m_{\text{Al}} = 500 \, \text{g}\) (mass of aluminum)
- \(c_{\text{Al}}\) is the specific heat of aluminum (to be determined)
- \(T_i = 100 \, \text{°C}\) (initial temperature of aluminum)
- \(T_f = 46.8 \, \text{°C}\) (final temperature)

### Step 6: Calculate the change in temperature for aluminum
The change in temperature for aluminum is:
\[ \Delta T = T_i - T_f = 100 \, \text{°C} - 46.8 \, \text{°C} = 53.2 \, \text{°C} \]

### Step 7: Set the heat gained equal to the heat lost
Equating the heat gained and heat lost:
\[ 5821.2 = 500 \cdot c_{\text{Al}} \cdot 53.2 \]

### Step 8: Solve for the specific heat of aluminum
Rearranging the equation to solve for \(c_{\text{Al}}\):
\[ c_{\text{Al}} = \frac{5821.2}{500 \cdot 53.2} \]
Calculating:
\[ c_{\text{Al}} = \frac{5821.2}{26600} \approx 0.22 \, \text{cal/g°C} \]

### Final Answer
The specific heat of aluminum is approximately \(0.22 \, \text{cal/g°C}\).

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