Let's solve this problem step by step.
Given:
- Mass of the bomb calorimeter ([tex]\(m\)[/tex]): [tex]\(1.20 \text{ kg} = 1200 \text{ g}\)[/tex] (since 1 kg = 1000 g)
- Specific heat capacity of the calorimeter ([tex]\(C_p\)[/tex]): [tex]\(3.55 \text{ J/g°C}\)[/tex]
- Initial temperature of the calorimeter ([tex]\(T_{\text{initial}}\)[/tex]): [tex]\(22.5^{\circ}\text{C}\)[/tex]
- Heat released during the combustion ([tex]\(q\)[/tex]): [tex]\(14.0 \text{ kJ} = 14000 \text{ J}\)[/tex] (since 1 kJ = 1000 J)
We need to find the final temperature ([tex]\(T_{\text{final}}\)[/tex]) of the calorimeter.
First, we recall the formula to find the change in temperature ([tex]\(\Delta T\)[/tex]):
[tex]\[ q = m \cdot C_p \cdot \Delta T \][/tex]
From this formula, we can solve for [tex]\(\Delta T\)[/tex]:
[tex]\[ \Delta T = \frac{q}{m \cdot C_p} \][/tex]
Plugging in the values:
[tex]\[ \Delta T = \frac{14000 \text{ J}}{1200 \text{ g} \cdot 3.55 \text{ J/g°C}} \][/tex]
Let's do the division:
[tex]\[ \Delta T = \frac{14000}{4260} \approx 3.29^{\circ}\text{C} \][/tex]
The change in temperature is approximately [tex]\(3.29^{\circ}\text{C}\)[/tex].
Now, we add this change in temperature to the initial temperature of the calorimeter to find the final temperature:
[tex]\[ T_{\text{final}} = T_{\text{initial}} + \Delta T \][/tex]
[tex]\[ T_{\text{final}} = 22.5^{\circ}\text{C} + 3.29^{\circ}\text{C} \][/tex]
[tex]\[ T_{\text{final}} \approx 25.79^{\circ}\text{C} \][/tex]
Rounding to one decimal place, the final temperature of the calorimeter is:
[tex]\[ T_{\text{final}} \approx 25.8^{\circ}\text{C} \][/tex]
Therefore, the correct answer is
[tex]\[ 25.8^{\circ} \text{C} \][/tex]
