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

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

Initial data: m1 (mass of sunflower oil) = 200 g; m2 (weight of a steel spoon) = 50 g; Δt2 (change in temperature of the steel spoon) = 20 ºС.

Constants: C1 (specific heat capacity of sunflower oil) = 1700 J / (kg * K); C2 (specific heat capacity of steel) = 500 J / (kg * K).

SI system: m1 = 200 g = 200/1000 kg = 0.2 kg; m2 = 50 g = 50/1000 kg = 0.05 kg.

We will assume that there is no heat loss and Q1 = Q2.

C1 * m1 * Δt1 = C2 * m2 * Δt2.

Δt1 = C2 * m2 * Δt2 / (C1 * m1) = 500 * 0.05 * 20 / (1700 * 0.2) ≈ 1.5 ºС.