To calculate the calorimeter constant, you can use the principle of conservation of energy. The heat lost by the hot water equals the heat gained by the cold water and the calorimeter.

The formula is:

$Q_{\text{lost by hot water}} = Q_{\text{gained by cold water}} + Q_{\text{gained by calorimeter}}$

Where:

• $Q = mc\Delta T$ (heat transferred, $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature).
• The specific heat capacity of water ($c$) is $4.18 \, \text{J/g°C}$.

We need to find the calorimeter constant ($C_{\text{cal}}$):

$Q_{\text{calorimeter}} = C_{\text{cal}} \cdot \Delta T_{\text{cal}}$

Step-by-step process:

1. Heat lost by hot water: $Q_{\text{hot}} = m_{\text{hot}} \cdot c_{\text{water}} \cdot (T_{\text{hot initial}} - T_{\text{final}})$ Substitute values:
   $m_{\text{hot}} = 90.0 \, \text{g}$,
   $T_{\text{hot initial}} = 89.0 \,°C$,
   $T_{\text{final}} = 43.0 \,°C$,
   $c_{\text{water}} = 4.18 \, \text{J/g°C}$.
   $Q_{\text{hot}} = 90.0 \cdot 4.18 \cdot (89.0 - 43.0) = 90.0 \cdot 4.18 \cdot 46.0 = 17327.4 \, \text{J}$

2. Heat gained by cold water:
   $Q_{\text{cold}} = m_{\text{cold}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{cold initial}})$ Substitute values:
   $m_{\text{cold}} = 99.0 \, \text{g}$,
   $T_{\text{cold initial}} = 3.0 \,°C$,
   $T_{\text{final}} = 43.0 \,°C$.
   $Q_{\text{cold}} = 99.0 \cdot 4.18 \cdot (43.0 - 3.0) = 99.0 \cdot 4.18 \cdot 40.0 = 16513.2 \, \text{J}$

3. Heat absorbed by the calorimeter: $Q_{\text{hot}} = Q_{\text{cold}} + Q_{\text{calorimeter}}$ Rearranging to find $Q_{\text{calorimeter}}$: $Q_{\text{calorimeter}} = Q_{\text{hot}} - Q_{\text{cold}} = 17327.4 - 16513.2 = 814.2 \, \text{J}$

4. Find the calorimeter constant:
   The temperature change of the calorimeter is the same as that of the water, which is: $\Delta T_{\text{cal}} = 43.0 - 3.0 = 40.0 \,°C$ Now, use: $C_{\text{cal}} = \frac{Q_{\text{calorimeter}}}{\Delta T_{\text{cal}}} = \frac{814.2}{40.0} = 20.36 \, \text{J/°C}$

Thus, the calorimeter constant is 20.36 J/°C.

Would you like more details or have any further questions?

Related Questions:

1. How does the specific heat capacity affect the heat transfer in this experiment?
2. Why is the final temperature of the mixture lower than the initial hot water temperature?
3. What assumptions are made in this calorimetry experiment?
4. How would the results change if the calorimeter had a higher heat capacity?
5. How can you reduce experimental error in calorimeter experiments?

Tip: Always ensure that the calorimeter is well-insulated to minimize heat exchange with the surroundings, which could affect accuracy.