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To calculate the calorimeter constant (\(C_{\text{cal}}\)), we typically use the heat transfer equation based on the principle of conservation of energy. The heat lost by the hot water is gained by the cold water and the calorimeter. The formula to determine the calorimeter constant is:

\[
\text{Heat lost by hot water} = \text{Heat gained by cold water} + \text{Heat absorbed by calorimeter}
\]

1. **Heat lost by hot water:**

\[
q_{\text{hot}} = m_{\text{hot}} \cdot c_{\text{water}} \cdot (T_{\text{hot initial}} - T_{\text{final}})
\]

2. **Heat gained by cold water:**

\[
q_{\text{cold}} = m_{\text{cold}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{cold initial}})
\]

3. **Heat absorbed by the calorimeter:**

\[
q_{\text{cal}} = C_{\text{cal}} \cdot (T_{\text{final}} - T_{\text{initial}})
\]

Here:

- \( m_{\text{hot}} \) = mass of hot water = 89.99 g
- \( m_{\text{cold}} \) = mass of cold water = 99.0 g
- \( c_{\text{water}} \) = specific heat capacity of water = 4.18 J/g°C
- \( T_{\text{hot initial}} \) = initial temperature of hot water = 89.0°C
- \( T_{\text{cold initial}} \) = initial temperature of cold water = 3.0°C
- \( T_{\text{final}} \) = final temperature after mixing = 43.0°C

Now, the heat balance equation becomes:

\[
m_{\text{hot}} \cdot c_{\text{water}} \cdot (T_{\text{hot initial}} - T_{\text{final}}) = m_{\text{cold}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{cold initial}}) + C_{\text{cal}} \cdot (T_{\text{final}} - T_{\text{initial}})
\]

### Step 1: Calculate the heat lost by hot water:
\[
q_{\text{hot}} = 89.99 \, \text{g} \times 4.18 \, \text{J/g°C} \times (89.0 - 43.0)
\]
\[
q_{\text{hot}} = 89.99 \times 4.18 \times 46 = 17294.57 \, \text{J}
\]

### Step 2: Calculate the heat gained by cold water:
\[
q_{\text{cold}} = 99.0 \, \text{g} \times 4.18 \, \text{J/g°C} \times (43.0 - 3.0)
\]
\[
q_{\text{cold}} = 99.0 \times 4.18 \times 40 = 16512.0 \, \text{J}
\]

### Step 3: Heat absorbed by the calorimeter:
\[
q_{\text{cal}} = q_{\text{hot}} - q_{\text{cold}} = 17294.57 - 16512.0 = 782.57 \, \text{J}
\]

Finally, the calorimeter constant is:
\[
C_{\text{cal}} = \frac{q_{\text{cal}}}{T_{\text{final}} - T_{\text{initial}}}
\]
\[
C_{\text{cal}} = \frac{782.57 \, \text{J}}{43.0 - 3.0} = \frac{782.57}{40} = 19.56 \, \text{J/°C}
\]

So, the calorimeter constant is approximately **19.56 J/°C**.

Would you like any further details or clarifications?

### Related Questions:
1. What assumptions are made in this heat transfer calculation?
2. How would the result change if the specific heat capacity of water was different?
3. What would happen if the calorimeter was not properly insulated?
4. How does the mass of the water affect the final temperature after mixing?
5. Can the calorimeter constant be calculated using different methods?

### Tip:
Always ensure that all temperatures and units are consistent when calculating heat transfer in calorimetry experiments.

Learn how to calculate the calorimeter constant with a detailed solution. The problem involves thermodynamics and heat transfer principles, using the specific heat of water and conservation of energy. Suitable for students in Grades 11-12.

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