To find the change in energy ($\Delta E$) for the combustion reaction in the calorimeter, we can use the following formula:

$q = C \cdot \Delta T$

where:

• $q$ is the heat absorbed or released (in joules, $J$),
• $C$ is the heat capacity of the calorimeter (in $J/^\circ C$),
• $\Delta T$ is the change in temperature (in $^\circ C$).

Given:

• Heat capacity $C = 2970 \, J/^\circ C$,
• Temperature increase $\Delta T = 4.50^\circ C$.

Let's substitute these values:

$q = 2970 \, J/^\circ C \times 4.50^\circ C$

Calculating:

$q = 2970 \times 4.50 = 13365 \, J$

Since the problem asks for the answer in kilojoules, we can convert $q$ to kilojoules by dividing by 1000:

$\Delta E = \frac{13365}{1000} = 13.365 \, \text{kJ}$

Answer:

The change in energy ($\Delta E$) is 13.365 kJ.

Would you like further clarification or a deeper explanation?

Related Questions:

1. How would the change in energy vary if the heat capacity of the calorimeter was doubled?
2. What would happen to the temperature change if only half the sample mass was burned?
3. How can we determine if this reaction is exothermic or endothermic based on the setup?
4. Why is it important that the calorimeter is a "bomb calorimeter" in this setup?
5. How does the combustion of hydrocarbons generally affect the energy output in such calorimetry experiments?

Tip:

Remember, in calorimetry problems, make sure to carefully distinguish between specific heat and heat capacity—they’re different properties that affect calculations differently.