The combustion of sucrose can be carried out in a calorimeter. If 343 mg of sucrose is combusted in a calorimeter containing 1.00 L of water, predict the change in temperature of the Calorimetry assembly (assume the calorimeter has the same specific heat capacity as water).

To predict the change in temperature of the calorimeter assembly during the combustion of sucrose, we can follow these steps:

### 1. Write the balanced chemical equation for the combustion of sucrose:
The chemical formula for sucrose is \( C_{12}H_{22}O_{11} \). The balanced combustion reaction in oxygen is:

\[
C_{12}H_{22}O_{11}(s) + 12O_2(g) \rightarrow 12CO_2(g) + 11H_2O(l)
\]

### 2. Determine the amount of heat released during the combustion:
The standard enthalpy change of combustion (\( \Delta H_{comb} \)) for sucrose is typically around \(-5645 \, \text{kJ/mol}\).

### 3. Convert the mass of sucrose to moles:
Given that 343 mg (0.343 g) of sucrose is combusted, calculate the number of moles:

\[
\text{Molar mass of sucrose} = 12(12.01) + 22(1.008) + 11(16.00) = 342.30 \, \text{g/mol}
\]

\[
\text{Moles of sucrose} = \frac{0.343 \, \text{g}}{342.30 \, \text{g/mol}} = 0.001002 \, \text{mol}
\]

### 4. Calculate the heat released:
Using the molar enthalpy of combustion:

\[
\text{Heat released} = 0.001002 \, \text{mol} \times (-5645 \, \text{kJ/mol}) = -5.66 \, \text{kJ}
\]

### 5. Relate the heat released to the temperature change:
The calorimeter contains 1.00 L of water, which is equivalent to 1.00 kg of water (since the density of water is approximately \(1.00 \, \text{g/mL}\)).

The specific heat capacity of water is \(4.18 \, \text{J/g}^\circ\text{C}\).

The heat change is related to temperature change (\(\Delta T\)) by the formula:

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

Where:
- \( q \) is the heat released (in joules),
- \( m \) is the mass of the water (in grams),
- \( c \) is the specific heat capacity (in \( \text{J/g}^\circ\text{C} \)),
- \( \Delta T \) is the change in temperature (in \( ^\circ\text{C} \)).

\[
5.66 \, \text{kJ} = 5660 \, \text{J}
\]

\[
\Delta T = \frac{5660 \, \text{J}}{1000 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C}} = \frac{5660 \, \text{J}}{4180 \, \text{J/}^\circ\text{C}} \approx 1.35^\circ\text{C}
\]

### **Conclusion:**
The predicted change in temperature of the calorimetry assembly is approximately **1.35°C**.

Do you want more details or have any questions?  

Here are 5 related questions for further exploration:
1. How would the change in temperature differ if the mass of sucrose was doubled?
2. What role does the specific heat capacity of water play in this calculation?
3. How would the temperature change if the calorimeter contained a different substance instead of water?
4. What are some potential sources of error in a real-life calorimetry experiment?
5. How does the heat capacity of the calorimeter itself affect the final temperature?

**Tip:** Always ensure that the calorimeter is properly insulated to minimize heat loss during combustion experiments.

Learn how to predict the change in temperature of a calorimeter assembly when 343 mg of sucrose is combusted in 1.00 L of water. This problem involves thermochemistry, heat transfer calculations, and specific heat capacity of water. Suitable for high school students studying chemistry.

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