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 solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the hot water will be equal to the heat gained by the cold water and the calorimeter.

### Step-by-Step Solution:

1. **Identify the known values:**
- 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 (T) = 28.7 °C

2. **Write the heat loss and heat gain equations:**
- Heat lost by hot water (Q_loss) = M2 * c * (T2 - T)
- Heat gained by cold water (Q_gain) = M1 * c * (T - T1)
- Heat gained by the calorimeter (Q_calorimeter) = Mw * c * (T - T1), where Mw is the water equivalent of the calorimeter.

Here, c (specific heat of water) can be taken as 1 (in appropriate units).

3. **Set up the equation based on conservation of energy:**
\[ Q_{loss} = Q_{gain} + Q_{calorimeter} \]
This can be expressed as:
\[ M2 \cdot (T2 - T) = M1 \cdot (T - T1) + Mw \cdot (T - T1) \]

4. **Substitute the known values into the equation:**
\[ 143.7 \cdot (36.5 - 28.7) = 70.2 \cdot (28.7 - 15.3) + Mw \cdot (28.7 - 15.3) \]

5. **Calculate the left side (heat lost by hot water):**
\[ 143.7 \cdot 7.8 = 1127.86 \text{ J} \]

6. **Calculate the right side (heat gained by cold water):**
\[ 70.2 \cdot 13.4 = 941.68 \text{ J} \]

7. **Combine the right side:**
\[ 1127.86 = 941.68 + Mw \cdot 13.4 \]

8. **Isolate Mw:**
\[ Mw \cdot 13.4 = 1127.86 - 941.68 \]
\[ Mw \cdot 13.4 = 186.18 \]
\[ Mw = \frac{186.18}{13.4} \approx 13.91 \text{ g} \]

### Final Answer:
The water equivalent of the calorimeter is approximately **13.91 g**.