The temperature of 380 ml water will change by how much if it receives all the energy released during the cooling

The temperature of 380 ml water will change by how much if it receives all the energy released during the cooling of a copper cylinder with a mass of 0.15 from a temperature of 90 degrees to a temperature of 20 degrees.

Data: V (volume of water) = 380 ml = 380 cm ^ 3 = 380 * 10 ^ -6 m ^ 3; m2 (cylinder mass) = 0.15 kg; t1 (initial cylinder temperature) = 90 ºС, t2 (end cylinder temperature) = 20 ºС.

Constants: ρ (water density) = 1000 kg / m ^ 3; C1 (specific heat capacity of water) = 4200 J / (kg * K); C2 (specific heat capacity of copper) = 400 J / (kg * K).

Equality: C1 * m1 * Δt = C2 * m2 * (t1 – t2).

C1 * V * ρ * Δt = C2 * m2 * (t1 – t2).

Δt = C2 * m2 * (t1 – t2) / (C1 * V * ρ) = 400 * 0.15 * (90 – 20) / (4200 * 380 * 10 ^ -6 * 1000) ≈ 2.6 ºС.