To find the enthalpy of the reaction in [tex]\( \text{kJ/g} \)[/tex], we need to follow a few steps. The main points to focus on are the heat capacity of the calorimeter, the initial and final temperatures, and the mass of the banana. Here’s a detailed step-by-step solution:
1. Determine the change in temperature (ΔT):
Given:
- Initial temperature, [tex]\( T_i = 20.2^\circ \text{C} \)[/tex]
- Final temperature, [tex]\( T_f = 22.0^\circ \text{C} \)[/tex]
The change in temperature, ΔT, is calculated as:
[tex]\[
\Delta T = T_f - T_i
\][/tex]
[tex]\[
\Delta T = 22.0^\circ \text{C} - 20.2^\circ \text{C} = 1.8^\circ \text{C}
\][/tex]
2. Calculate the heat absorbed by the calorimeter (q):
Given:
- Heat capacity of the calorimeter, [tex]\( C = 10.3 \text{ kJ/}^\circ \text{C} \)[/tex]
The heat absorbed, [tex]\( q \)[/tex], is given by:
[tex]\[
q = C \times \Delta T
\][/tex]
[tex]\[
q = 10.3 \text{ kJ/}^\circ \text{C} \times 1.8^\circ \text{C} = 18.54 \text{ kJ}
\][/tex]
3. Calculate the enthalpy of the reaction (ΔH) per gram of banana:
Given:
- Mass of the banana, [tex]\( m = 5.00 \text{ g} \)[/tex]
The enthalpy change, [tex]\( \Delta H \)[/tex], per gram of banana is calculated by:
[tex]\[
\Delta H = \frac{q}{\text{mass of the banana}}
\][/tex]
[tex]\[
\Delta H = \frac{18.54 \text{ kJ}}{5.00 \text{ g}} = 3.708 \text{ kJ/g}
\][/tex]
Hence, the enthalpy of the reaction is:
[tex]\[
\Delta H = +3.708 \text{ kJ/g}
\][/tex]
Note the significant figures are preserved throughout the calculation, reflecting the precision of the given data.
