A heated steel spoon weighing 50 g is lowered into sunflower oil weighing 200 g.

A heated steel spoon weighing 50 g is lowered into sunflower oil weighing 200 g. The spoon cools down by 20 degrees. How hot is the oil?

Q1 = Q2.

Q1 (cooling the steel spoon); Q1 = C1 * m1 * Δt1, where C is the specific heat capacity (C1 = 500 J / (K * kg)), m1 is the weight of the spoon (m1 = 50 g = 0.05 kg), Δt1 is the change in the temperature of the spoon (Δt1 = 20 ° C).

Q2 (heating of sunflower oil); Q2 = C2 * m2 * Δt2, where C2 is the specific heat capacity (C2 = 1700 J / (K * kg)), m2 is the mass of oil (m2 = 200 g = 0.2 kg), Δt2 is the change in oil temperature.

C1 * m1 * Δt1 = C2 * m2 * Δt2.

500 * 0.05 * 20 = 1700 * 0.2 * Δt2.

Δt2 = 500 * 0.05 * 20 / (1700 * 0.2) = 1.47 ° C.