Question

How much energy is released by the reaction if the temperature of the calorimeter rises by 2.87°C?

Answer 1

I don’t have the numbers for the calorimeter/solution or the amount of reactant, so I can’t give a single numeric answer. Here’s how to get it and a short example.

General method and sign convention
- Heat absorbed by calorimeter/solution: qabsorbed = (m·c + Ccal) · ΔT
- m = mass of solution (g), c = specific heat (J/g·°C), Ccal = heat capacity of the calorimeter (J/°C). If you have no separate calorimeter heat capacity, set Ccal = 0.
- Temperature change ΔT = Tfinal − Tinitial (use +2.87°C here).
- Heat released by the reaction: qrxn = − qabsorbed (reaction heat is the negative of heat absorbed by the calorimeter/solution).
- Convert to kJ if desired (1 kJ = 1000 J) and divide by moles of reactant to get kJ per mole.

Example (coffee‑cup calorimeter, assume 100.0 g solution, c = 4.184 J/g·°C, no calorimeter heat capacity)
- qabsorbed = m·c·ΔT = 100.0 g × 4.184 J/g·°C × 2.87 °C = 1.2008 × 10^3 J = 1.2008 kJ
- qrxn = −1.2008 kJ (the reaction released 1.20 kJ)

If you give me the mass of solution, its specific heat (or assume water = 4.184 J/g·°C), the calorimeter heat capacity (if any), and the amount (moles) of reactant, I’ll compute the exact energy released.