We want to solve Q = mcΔT for the liquid water; its change in temperature will tell us the amount of thermal energy that flowed out of the reaction. The specific heat, c, of water is 4.184 J/g °C.
Q = (72.0 g)(4.184 J/g °C)(100 °C - 25 °C) = 22593.6 J
Q ≈ 2.26 × 10⁴ J or 22.6 kJ (three significant figures).
