In an experiment to determine the specific heat of a metal,` a 0.20 kg` block of the metal at `150 .^(@) C` is dropped in a copper calorimeter (of water equivalent `0.025 kg` containing `150 cm^3` of water at `27 .^(@) C`. The final temperature is `40.^(@) C`. The specific heat of the metal is.

Options

A. `(a)0.1 Jg^-1 .^(@) C^-1`.
B. `(b)0.2 Jg^-1 .^(@) C^-1`.
C. `(c)0.3 cal g^-1 .^(@) C^-1`.
D. `(d)0.1 cal g^-1 .^(@) C^-1`.

Explanation

To determine the specific heat of the metal, we will follow these steps:

### Step 1: Identify the heat lost by the metal
The heat lost by the metal can be calculated using the formula:
\[ Q_{\text{lost}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot \Delta T_{\text{metal}} \]
Where:
- \( m_{\text{metal}} = 0.20 \, \text{kg} = 200 \, \text{g} \)
- \( c_{\text{metal}} \) is the specific heat of the metal (unknown)
- \( \Delta T_{\text{metal}} = T_{\text{initial, metal}} - T_{\text{final}} = 150^\circ C - 40^\circ C = 110^\circ C \)

Thus, the heat lost by the metal is:
\[ Q_{\text{lost}} = 200 \cdot c_{\text{metal}} \cdot 110 \]

### Step 2: Identify the heat gained by the water and calorimeter
The heat gained by the water and calorimeter can be calculated as:
\[ Q_{\text{gained}} = Q_{\text{water}} + Q_{\text{calorimeter}} \]
Where:
- For water:
- Volume of water = 150 cm³
- Mass of water = 150 g (since density of water = 1 g/cm³)
- Specific heat of water \( c_{\text{water}} = 1 \, \text{cal/g}^\circ C \)
- Change in temperature \( \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 40^\circ C - 27^\circ C = 13^\circ C \)

Thus, the heat gained by the water is:
\[ Q_{\text{water}} = 150 \cdot 1 \cdot 13 = 1950 \, \text{cal} \]

- For the calorimeter (water equivalent):
- Water equivalent of the calorimeter = 0.025 kg = 25 g
- Specific heat of calorimeter \( c_{\text{calorimeter}} = 1 \, \text{cal/g}^\circ C \)
- Change in temperature \( \Delta T_{\text{calorimeter}} = 40^\circ C - 27^\circ C = 13^\circ C \)

Thus, the heat gained by the calorimeter is:
\[ Q_{\text{calorimeter}} = 25 \cdot 1 \cdot 13 = 325 \, \text{cal} \]

### Step 3: Calculate total heat gained
Now, we can find the total heat gained:
\[ Q_{\text{gained}} = Q_{\text{water}} + Q_{\text{calorimeter}} = 1950 + 325 = 2275 \, \text{cal} \]

### Step 4: Set heat lost equal to heat gained
According to the principle of conservation of energy:
\[ Q_{\text{lost}} = Q_{\text{gained}} \]
Thus:
\[ 200 \cdot c_{\text{metal}} \cdot 110 = 2275 \]

### Step 5: Solve for the specific heat of the metal
Now, we can solve for \( c_{\text{metal}} \):
\[ c_{\text{metal}} = \frac{2275}{200 \cdot 110} \]
Calculating the denominator:
\[ 200 \cdot 110 = 22000 \]
Thus:
\[ c_{\text{metal}} = \frac{2275}{22000} \approx 0.1034 \, \text{cal/g}^\circ C \]

### Final Answer
The specific heat of the metal is approximately:
\[ c_{\text{metal}} \approx 0.1 \, \text{cal/g}^\circ C \]

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