A calorimeter contains 10 g of water at `20^(@)C`. The temperature falls to `15^(@)C` in 10 min. When calorimeter contains 20 g of water at `20^(@)C`, it takes 15 min for the temperature to becomes `15^(@)C`. The water equivalent of the calorimeter is

Options

A. 5 g
B. 10 g
C. 25 g
D. 50 g

Explanation

To find the water equivalent of the calorimeter, we can use the principle of heat transfer. The heat lost by the water will be equal to the heat gained by the calorimeter (including its water equivalent).

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( m_1 = 10 \, \text{g} \) (mass of water in the first case)
- Let \( m_2 = 20 \, \text{g} \) (mass of water in the second case)
- Let \( W \) be the water equivalent of the calorimeter.
- The specific heat of water, \( S = 1 \, \text{cal/g°C} \).
- The initial temperature \( T_i = 20 \, \text{°C} \) and final temperature \( T_f = 15 \, \text{°C} \).

2. **Calculate the Heat Lost in the First Case**:
- The change in temperature \( \Delta T_1 = T_f - T_i = 15 - 20 = -5 \, \text{°C} \).
- Heat lost by the water in the first case:
\[ Q_1 = m_1 \cdot S \cdot \Delta T_1 = 10 \cdot 1 \cdot (-5) = -50 \, \text{cal} \]
- The heat lost by the calorimeter (including its water equivalent) can be expressed as:
\[ Q_1 = W \cdot \Delta T_1 = W \cdot (-5) \]

3. **Set Up the Equation for the First Case**:
- Since the heat lost by the water equals the heat gained by the calorimeter:
\[ -50 = W \cdot (-5) \quad \Rightarrow \quad 50 = 5W \quad \Rightarrow \quad W = 10 \, \text{g} \]

4. **Calculate the Heat Lost in the Second Case**:
- The change in temperature \( \Delta T_2 = T_f - T_i = 15 - 20 = -5 \, \text{°C} \).
- Heat lost by the water in the second case:
\[ Q_2 = m_2 \cdot S \cdot \Delta T_2 = 20 \cdot 1 \cdot (-5) = -100 \, \text{cal} \]
- The heat lost by the calorimeter (including its water equivalent) can be expressed as:
\[ Q_2 = W \cdot \Delta T_2 = W \cdot (-5) \]

5. **Set Up the Equation for the Second Case**:
- Since the heat lost by the water equals the heat gained by the calorimeter:
\[ -100 = W \cdot (-5) \quad \Rightarrow \quad 100 = 5W \quad \Rightarrow \quad W = 20 \, \text{g} \]

6. **Equate the Two Cases**:
- From both cases, we have:
\[ 50 = 5W \quad \text{and} \quad 100 = 5W \]
- Solving these equations gives us the same value of \( W \).

7. **Final Result**:
- The water equivalent of the calorimeter is \( W = 10 \, \text{g} \).

### Conclusion:
The water equivalent of the calorimeter is **10 g**.