A calorimeter contains `50 g` of water at `50^(circ) C`. The temperature falls to `45^(circ) C` in `10 min`. When the calorimeter contain `100 g` of water at `cdot 50^(circ) C`, it takes 15 min' for the temperature to become `45^(circ) C`. Find the water equivalent (in g) of the calorimeter. (Assume Newton's law of cooling)

Answer

To solve the problem of finding the water equivalent of the calorimeter using Newton's law of cooling, we can follow these steps:

### Step 1: Define the given values
- For the first case:
- Mass of water, \( m_1 = 50 \, \text{g} = 0.05 \, \text{kg} \)
- Initial temperature, \( T_{i1} = 50^\circ C \)
- Final temperature, \( T_{f1} = 45^\circ C \)
- Time taken, \( t_1 = 10 \, \text{min} = 600 \, \text{s} \)

- For the second case:
- Mass of water, \( m_2 = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Initial temperature, \( T_{i2} = 50^\circ C \)
- Final temperature, \( T_{f2} = 45^\circ C \)
- Time taken, \( t_2 = 15 \, \text{min} = 900 \, \text{s} \)

### Step 2: Calculate the change in temperature
- For both cases, the change in temperature \( \Delta T \) is:
\[ \Delta T = T_{i} - T_{f} = 50^\circ C - 45^\circ C = 5^\circ C \]

### Step 3: Apply Newton's law of cooling
According to Newton's law of cooling, the heat lost by the water and the calorimeter can be expressed as:
\[ Q = (m + W) c \Delta T \]
where \( W \) is the water equivalent of the calorimeter and \( c \) is the specific heat capacity of water (approximately \( 4200 \, \text{J/kg}^\circ C \)).

### Step 4: Set up the equations for both cases
1. For the first case:
\[ \frac{Q_1}{t_1} = (m_1 + W) c \Delta T / t_1 \]
\[ \frac{Q_1}{600} = (0.05 + W) \cdot 4200 \cdot 5 / 600 \]

2. For the second case:
\[ \frac{Q_2}{t_2} = (m_2 + W) c \Delta T / t_2 \]
\[ \frac{Q_2}{900} = (0.1 + W) \cdot 4200 \cdot 5 / 900 \]

### Step 5: Simplify both equations
1. From the first case:
\[ \frac{(0.05 + W) \cdot 4200 \cdot 5}{600} = \frac{Q_1}{600} \]
\[ 4200 \cdot 5 \cdot (0.05 + W) = Q_1 \]

2. From the second case:
\[ \frac{(0.1 + W) \cdot 4200 \cdot 5}{900} = \frac{Q_2}{900} \]
\[ 4200 \cdot 5 \cdot (0.1 + W) = Q_2 \]

### Step 6: Equate the heat loss expressions
Since both \( Q_1 \) and \( Q_2 \) represent the same heat loss due to cooling, we can set the two expressions equal:
\[ 4200 \cdot 5 \cdot (0.05 + W) / 600 = 4200 \cdot 5 \cdot (0.1 + W) / 900 \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 900(0.05 + W) = 600(0.1 + W) \]
Expanding both sides:
\[ 45 + 900W = 60 + 600W \]
Rearranging gives:
\[ 900W - 600W = 60 - 45 \]
\[ 300W = 15 \]
\[ W = \frac{15}{300} = 0.05 \, \text{kg} = 50 \, \text{g} \]

### Final Answer
The water equivalent of the calorimeter is \( 50 \, \text{g} \).

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