A calorimeter contains 70.2 g of water at `15.3^@C`. IF 143.7 g of water at `36.5^@C` is mixed with it, the common temperature becomes `28.7^@C`. The water equivalent of a calorimeter is

Options

A. 15.6 g
B. 9.4 g
C. 6.3 g
D. 13.4 g

Answer

To find the water equivalent of the calorimeter, we can use the principle of calorimetry, which states that the heat lost by the hot water will be equal to the heat gained by the cold water and the calorimeter itself.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of cold water (m1) = 70.2 g
- Initial temperature of cold water (T1) = 15.3 °C
- Mass of hot water (m2) = 143.7 g
- Initial temperature of hot water (T2) = 36.5 °C
- Final common temperature (Tf) = 28.7 °C

2. **Set Up the Heat Transfer Equation:**
The heat gained by the cold water and the calorimeter is equal to the heat lost by the hot water:
\[ \text{Heat gained by cold water + Heat gained by calorimeter} = \text{Heat lost by hot water} \]

This can be expressed mathematically as:
\[ (m_1 + m) \cdot c \cdot (T_f - T_1) + m \cdot c \cdot (T_f - T_1) = m_2 \cdot c \cdot (T_2 - T_f) \]
where \( m \) is the water equivalent of the calorimeter, and \( c \) is the specific heat capacity of water (which cancels out).

3. **Substitute the Values:**
Since \( c \) cancels out, we can simplify the equation:
\[ (70.2 + m)(28.7 - 15.3) + m(28.7 - 15.3) = 143.7(36.5 - 28.7) \]

Simplifying the temperature differences:
\[ (70.2 + m)(13.4) + m(13.4) = 143.7(7.8) \]

4. **Combine Like Terms:**
\[ (70.2 + m + m)(13.4) = 143.7 \cdot 7.8 \]
\[ (70.2 + 2m)(13.4) = 1127.86 \]

5. **Solve for m:**
Expanding the left side:
\[ 70.2 \cdot 13.4 + 2m \cdot 13.4 = 1127.86 \]
\[ 941.68 + 26.8m = 1127.86 \]
\[ 26.8m = 1127.86 - 941.68 \]
\[ 26.8m = 186.18 \]
\[ m = \frac{186.18}{26.8} \approx 6.94 \text{ g} \]

6. **Final Answer:**
The water equivalent of the calorimeter is approximately **6.94 g**.