To determine the enthalpy of the reaction per gram ([tex]\(\Delta H\)[/tex]) for the given problem, we will follow these steps:
1. Calculate the temperature change ([tex]\(\Delta T\)[/tex]):
[tex]\[
\Delta T = T_\text{final} - T_\text{initial}
\][/tex]
Given:
[tex]\[
T_\text{initial} = 20.2^\circ C
\][/tex]
[tex]\[
T_\text{final} = 22.0^\circ C
\][/tex]
[tex]\[
\Delta T = 22.0^\circ C - 20.2^\circ C = 1.8^\circ C
\][/tex]
2. Calculate the heat absorbed by the calorimeter ([tex]\(q\)[/tex]):
[tex]\[
q = \text{Heat Capacity} \times \Delta T
\][/tex]
Given:
[tex]\[
\text{Heat Capacity} = 10.3 \, \text{kJ/}^\circ \text{C}
\][/tex]
[tex]\[
q = 10.3 \, \text{kJ/}^\circ \text{C} \times 1.8^\circ C = 18.54 \, \text{kJ}
\][/tex]
3. Calculate the enthalpy change ([tex]\(\Delta H\)[/tex]) per gram of banana:
[tex]\[
\Delta H = \frac{q}{\text{mass of banana}}
\][/tex]
Given:
[tex]\[
\text{mass of banana} = 5.00 \, \text{g}
\][/tex]
[tex]\[
\Delta H = \frac{18.54 \, \text{kJ}}{5.00 \, \text{g}} = 3.708 \, \text{kJ/g}
\][/tex]
Therefore, the enthalpy of the reaction, [tex]\(\Delta H\)[/tex], is:
[tex]\[
\Delta H = +3.708 \, \text{kJ/g}
\][/tex]
This result indicates that [tex]\(3.708 \, \text{kJ}\)[/tex] of energy is absorbed per gram of banana during the reaction. Notice the positive sign indicates an endothermic reaction, meaning the system absorbs heat.
