We can use the equation:
q = m * c * ΔT
where q is the heat absorbed or released by the water, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.
Since we know that the calorimeter contains 100 mL (or 100 g, since 1 mL of water has a mass of 1 g) of water and that the temperature of the water increased by 9.3°C, we can plug in these values:
q = (100 g) * (4.184 J/g-°C) * (9.3°C)
q = 3896.68 J
However, this is not the total amount of heat produced by the reaction. We need to take into account the heat absorbed by the calorimeter itself, which has a heat capacity of 50.2 J/°C. If we assume that the temperature of the calorimeter did not change during the reaction (i.e., it remained constant), we can calculate the heat absorbed by the calorimeter:
q_calorimeter = (50.2 J/°C) * (9.3°C)
q_calorimeter = 466.86 J
The total heat produced by the reaction is then:
q_reaction = q_water + q_calorimeter
q_reaction = 3896.68 J + 466.86 J
q_reaction = 4363.54 J
Therefore, the heat produced by the reaction is 4363.54 J.