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To calculate the **calorimeter constant** (denoted as \( C_{\text{cal}} \) in J/°C), we need to use the principle of conservation of energy. The heat lost by the hot water is equal to the heat gained by the cold water plus the heat absorbed by the calorimeter.

Here's the formula we use:

\[
Q_{\text{hot}} = Q_{\text{cold}} + Q_{\text{calorimeter}}
\]

Where:
- \( Q_{\text{hot}} \) = heat lost by the hot water
- \( Q_{\text{cold}} \) = heat gained by the cold water
- \( Q_{\text{calorimeter}} \) = heat absorbed by the calorimeter, which is \( C_{\text{cal}} \times \Delta T \)

To calculate the heat for the water components, we use:

\[
Q = m \times c \times \Delta T
\]

Where:
- \( m \) = mass of water (g)
- \( c \) = specific heat capacity of water (\( 4.18 \, \text{J/g°C} \))
- \( \Delta T \) = change in temperature

### 1. Heat lost by the hot water:
\[
Q_{\text{hot}} = m_{\text{hot}} \times c_{\text{water}} \times (T_{\text{initial, hot}} - T_{\text{final}})
\]
Substitute values:
\[
Q_{\text{hot}} = 89.5 \times 4.18 \times (89.0 - 43.0)
\]

### 2. Heat gained by the cold water:
\[
Q_{\text{cold}} = m_{\text{cold}} \times c_{\text{water}} \times (T_{\text{final}} - T_{\text{initial, cold}})
\]
Substitute values:
\[
Q_{\text{cold}} = 99.0 \times 4.18 \times (43.0 - 3.0)
\]

### 3. Heat absorbed by the calorimeter:
\[
Q_{\text{calorimeter}} = C_{\text{cal}} \times (T_{\text{final}} - T_{\text{initial}})
\]
Substitute values and solve for \( C_{\text{cal}} \):
\[
C_{\text{cal}} = \frac{Q_{\text{hot}} - Q_{\text{cold}}}{(T_{\text{final}} - T_{\text{initial}})}
\]

Now, let's calculate each component:

1. \( Q_{\text{hot}} = 89.5 \times 4.18 \times (89.0 - 43.0) = 89.5 \times 4.18 \times 46 = 17185.18 \, \text{J} \)
2. \( Q_{\text{cold}} = 99.0 \times 4.18 \times (43.0 - 3.0) = 99.0 \times 4.18 \times 40 = 16512 \, \text{J} \)

Finally, solve for \( C_{\text{cal}} \):

\[
C_{\text{cal}} = \frac{17185.18 - 16512}{43.0 - 3.0} = \frac{673.18}{40} = 16.83 \, \text{J/°C}
\]

So, the calorimeter constant is approximately **16.83 J/°C**.

Let me know if you need more details or have any questions! 

Here are 5 related questions to explore further:
1. How would changing the masses of the water samples affect the calorimeter constant?
2. What assumptions are made in this type of calorimetry calculation?
3. How would we account for heat losses to the environment in this experiment?
4. Can this calorimeter constant be used for different liquids other than water?
5. How would a higher final temperature impact the calculated calorimeter constant?

**Tip:** Always ensure the specific heat capacity used matches the substance being measured (e.g., different liquids have different specific heat capacities).

Learn how to calculate the calorimeter constant using data from an experiment. This problem involves key thermodynamic concepts such as calorimetry and the conservation of energy. Suitable for students in Grades 10-12.

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