How many degrees can 510 g of copper be heated when using the same heat released during

How many degrees can 510 g of copper be heated when using the same heat released during cooling 2 liters of oil at 40 ° C

Given: m1 (mass of copper) = 510 g (0.51 kg); V2 (oil volume) = 2 l (2 * 10-3 m3); Δt2 (change in oil temperature) = 40 ºС.

Constants: C1 (specific heat of copper) = 400 J / (kg * K); ρ (density of sunflower oil) = 930 kg / m3; C2 (specific heat of vegetable oil) = 1700 J / (kg * ºС).

We express the change in copper temperature from the ratio: C1 * m1 * Δt1 = C2 * ρ * V2 * Δt2 = Q, whence Δt1 = C2 * ρ * V2 * Δt2 / (C1 * m1).

Calculation: Δt1 = 1700 * 930 * 2 * 10-3 * 40 / (400 * 0.51) = 620 ºС.