How many degrees will 510 g of copper heat up if it absorbs the same amount

How many degrees will 510 g of copper heat up if it absorbs the same amount of heat that is released when 2 liters of vegetable oil are cooled from 60 to 40ºC.

Given: m1 (copper) = 510 g or 0.51 kg; V2 (oil volume) = 2 l or 0.002 m3; t2н (start. oil temp.) = 60 ºС; t2к (final oil temp.) = 40 ºС.

Constants: C1 (heat capacity of copper) = 400 J / (kg * ºС); ρ2 (density of sunflower oil, 40 ºС) = 903 kg / m3; C2 (heat capacity of vegetable oil) = 1700 J / (kg * ºС).

The change in copper temperature is determined from the equality: C1 * m1 * Δt1 = C2 * m2 * (t2n – t2k), whence Δt1 = C2 * ρ2 * V2 * (t2n – t2k) / (C1 * m1) = 1700 * 903 * 0.002 * (60 – 40) / (400 * 0.51) = 301 ºС.