How much will the temperature of 380 ml water change if it receives all the energy released when cooling

How much will the temperature of 380 ml water change if it receives all the energy released when cooling a copper cylinder weighing 0.15 kg from a temperature of 90C to 20C

Let us express the temperature change of 380 ml of water from the equality: Sv * ρw * V * Δt = Cm * m2 * (t2 – t1), whence Δt = Cm * m2 * (t2 – t1) / (Sv * ρw * V), where Cm (specific heat capacity of copper) = 400 J / (kg * K); m2 (mass of the copper cylinder) = 0.15 kg; t2 (initial temperature) = 90 ºС; t1 (end temperature) = 20 ºС; Sv (specific heat capacity of water) = 4200 J / (kg * K); ρw (density of heated water) = 1000 kg / m3; V (water volume) = 380 ml (0.38 * 10-3 m3).
Δt = 400 * 0.15 * (90 – 20) / (4200 * 1000 * 0.80 * 10-3) = 2.63 ºС.