`100 ml` of water at `20^(@)C` and `100 ml` of water at `40^(@)C` are mixed in calorimeter until constant temperature reached. Now temperature of the mixture is `28^(@)C`. Water equivalent of calorimeter is

Options

A. 50 J
B. 104.5 J
C. `-24.2 J`
D. `209 J`

Explanation

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the hotter water will be equal to the heat gained by the cooler water and the calorimeter.

### Step-by-Step Solution

1. **Identify the Initial Conditions**:
- Volume of water at 20°C (V1) = 100 ml
- Volume of water at 40°C (V2) = 100 ml
- Final temperature of the mixture (T_f) = 28°C

2. **Calculate Heat Lost by Hot Water**:
- The heat lost by the hot water (Q_hot) can be calculated using the formula:
\[ Q_{\text{hot}} = m \cdot c \cdot (T_{\text{initial, hot}} - T_f) \]
- Here, \(m\) is the mass of water, \(c\) is the specific heat capacity of water (approximately 1 cal/g°C or 4.18 J/g°C), and \(T_{\text{initial, hot}} = 40°C\).
- Since the density of water is approximately 1 g/ml, the mass of 100 ml of water is 100 g.
- Therefore:
\[ Q_{\text{hot}} = 100 \cdot 4.18 \cdot (40 - 28) = 100 \cdot 4.18 \cdot 12 = 5016 \text{ J} \]

3. **Calculate Heat Gained by Cold Water**:
- The heat gained by the cold water (Q_cold) can be calculated using the formula:
\[ Q_{\text{cold}} = m \cdot c \cdot (T_f - T_{\text{initial, cold}}) \]
- Here, \(T_{\text{initial, cold}} = 20°C\).
- Therefore:
\[ Q_{\text{cold}} = 100 \cdot 4.18 \cdot (28 - 20) = 100 \cdot 4.18 \cdot 8 = 3344 \text{ J} \]

4. **Set Up the Equation for Heat Balance**:
- According to the principle of conservation of energy:
\[ Q_{\text{hot}} = Q_{\text{cold}} + Q_{\text{calorimeter}} \]
- Rearranging gives:
\[ Q_{\text{calorimeter}} = Q_{\text{hot}} - Q_{\text{cold}} = 5016 - 3344 = 1672 \text{ J} \]

5. **Calculate Water Equivalent of Calorimeter**:
- The water equivalent (W) of the calorimeter can be calculated using the formula:
\[ Q_{\text{calorimeter}} = W \cdot c \cdot (T_f - T_{\text{initial, cold}}) \]
- Rearranging gives:
\[ W = \frac{Q_{\text{calorimeter}}}{c \cdot (T_f - T_{\text{initial, cold}})} \]
- Substituting the known values:
\[ W = \frac{1672}{4.18 \cdot (28 - 20)} = \frac{1672}{4.18 \cdot 8} = \frac{1672}{33.44} \approx 50 \text{ g} \]

### Final Answer:
The water equivalent of the calorimeter is approximately **50 J**.