A calorimeter containes 50g of water at `50^(@)C` . The temperature falls to `45^(@)C` in 10 minutes. When the calorimeter contains 100g of water at `50^(@)C` it takes 18 minutes for the temperature to become `45^(@)C` . Find the water equivalent of the calorimeter.

Explanation

To solve the problem of finding the water equivalent of the calorimeter, we will apply the principle of heat transfer and the concept of calorimetry. We will use the data provided for two different scenarios involving the calorimeter.

### Step-by-Step Solution:

1. **Identify Given Data:**
- For the first case:
- Mass of water (m1) = 50 g = 0.05 kg
- Initial temperature (T_initial) = 50°C
- Final temperature (T_final) = 45°C
- Time (t1) = 10 minutes = 600 seconds
- For the second case:
- Mass of water (m2) = 100 g = 0.1 kg
- Time (t2) = 18 minutes = 1080 seconds

2. **Calculate the Temperature Change:**
- The change in temperature (ΔT) = T_initial - T_final = 50°C - 45°C = 5°C

3. **Set Up Heat Transfer Equations:**
- The heat lost by the water in both cases can be expressed as:
- For the first case (Q1):
\[ Q_1 = (m_1 + W) \cdot c \cdot \Delta T \]
Where \(W\) is the water equivalent of the calorimeter and \(c\) is the specific heat of water (approximately 4200 J/kg°C).
\[ Q_1 = (0.05 + W) \cdot 4200 \cdot 5 \]
- For the second case (Q2):
\[ Q_2 = (m_2 + W) \cdot c \cdot \Delta T \]
\[ Q_2 = (0.1 + W) \cdot 4200 \cdot 5 \]

4. **Calculate the Rate of Heat Transfer:**
- The rate of heat transfer for both cases can be expressed as:
- For the first case:
\[ \text{Rate}_1 = \frac{Q_1}{t_1} = \frac{(0.05 + W) \cdot 4200 \cdot 5}{600} \]
- For the second case:
\[ \text{Rate}_2 = \frac{Q_2}{t_2} = \frac{(0.1 + W) \cdot 4200 \cdot 5}{1080} \]

5. **Set the Rates Equal:**
- According to the principle of calorimetry, the rates of heat loss should be equal:
\[ \frac{(0.05 + W) \cdot 4200 \cdot 5}{600} = \frac{(0.1 + W) \cdot 4200 \cdot 5}{1080} \]

6. **Simplify the Equation:**
- Cancel out the common terms (4200 and 5):
\[ \frac{0.05 + W}{600} = \frac{0.1 + W}{1080} \]
- Cross-multiply to eliminate the fractions:
\[ 1080(0.05 + W) = 600(0.1 + W) \]

7. **Expand and Rearrange:**
- Expanding both sides:
\[ 54 + 1080W = 60 + 600W \]
- Rearranging gives:
\[ 1080W - 600W = 60 - 54 \]
\[ 480W = 6 \]
- Solving for W:
\[ W = \frac{6}{480} = 0.0125 \text{ kg} = 12.5 \text{ g} \]

### Final Answer:
The water equivalent of the calorimeter is **12.5 g**.