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, we need to find the water equivalent of the calorimeter based on the given conditions. Let's break it down step by step.

### Step 1: Understand the Given Data
- In the first case:
- Mass of water (m1) = 50 g
- Initial temperature (T1_initial) = 50°C
- Final temperature (T1_final) = 45°C
- Temperature change (ΔT1) = T1_initial - T1_final = 5°C
- Time taken (t1) = 10 minutes

- In the second case:
- Mass of water (m2) = 100 g
- Initial temperature (T2_initial) = 50°C
- Final temperature (T2_final) = 45°C
- Temperature change (ΔT2) = T2_initial - T2_final = 5°C
- Time taken (t2) = 18 minutes

### Step 2: Set Up the Heat Transfer Equations
Using the formula for heat transfer, we can express the heat lost by the water and the calorimeter in both cases.

For the first case:
\[ Q_1 = (m_1 + W) \cdot s \cdot \Delta T_1 \]
Where:
- \(W\) = water equivalent of the calorimeter
- \(s\) = specific heat of water (approximately 1 cal/g°C)

Thus, we have:
\[ Q_1 = (50 + W) \cdot 1 \cdot 5 \]
\[ Q_1 = (50 + W) \cdot 5 \]

For the second case:
\[ Q_2 = (m_2 + W) \cdot s \cdot \Delta T_2 \]
Thus, we have:
\[ Q_2 = (100 + W) \cdot 1 \cdot 5 \]
\[ Q_2 = (100 + W) \cdot 5 \]

### Step 3: Relate Heat Transfer to Time
The rate of heat loss can be expressed as:
\[ \frac{Q_1}{t_1} = \frac{Q_2}{t_2} \]

Substituting the expressions for \(Q_1\) and \(Q_2\):
\[ \frac{(50 + W) \cdot 5}{10} = \frac{(100 + W) \cdot 5}{18} \]

### Step 4: Simplify the Equation
We can cancel the 5 from both sides:
\[ \frac{50 + W}{10} = \frac{100 + W}{18} \]

Cross-multiplying gives:
\[ 18(50 + W) = 10(100 + W) \]

### Step 5: Expand and Rearrange
Expanding both sides:
\[ 900 + 18W = 1000 + 10W \]

Rearranging gives:
\[ 18W - 10W = 1000 - 900 \]
\[ 8W = 100 \]

### Step 6: Solve for W
Dividing both sides by 8:
\[ W = \frac{100}{8} = 12.5 \text{ g} \]

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

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