How much will the volume of water V1 = 380 ml change if it receives all the energy released during

How much will the volume of water V1 = 380 ml change if it receives all the energy released during the cooling of a copper cylinder with a mass of m2 = 0.15 kg. T1 = 90 ° C to t2 = 20 ° C

Data: V1 (volume of water taken) = 380 ml (380 * 10-6 m3, corresponding to m1 = 0.38 kg); m2 (mass of the cooled cylinder) = 0.15 kg; t1 (initial cylinder temperature) = 90 ºС; t2 (end temperature) = 20 ºС.

Constants: Sv (specific heat capacity of water) = 4200 J / (kg * K); Cm (specific heat capacity of copper) = 400 J / (kg * K).

Let us express the change in temperature of 380 ml of water from the equality: Cw * m1 * Δt = Cm * m2 * (t1 – t2); Δt = Cm * m2 * (t1 – t2) / (Cw * ρ * V1).

Calculation: Δt = 400 * 0.15 * (90 – 20) / (4200 * 0.38) = 2.63 ºС.