A 200 g steel bar heated in a furnace was dipped into 250 g water at a temperature of 15 ° C.

A 200 g steel bar heated in a furnace was dipped into 250 g water at a temperature of 15 ° C. The water temperature rose to 25 ° C. Calculate the oven temperature.

Q1 = Q2.

Q1 (cooling of a steel bar) = C * m1 * (t – t1), where C is beats. heat capacity of steel (assumed C = 500 J / (K * kg)), m1 – mass (m1 = 200 g = 0.2 kg), t – temp. in the oven, t1 – temp. equilibrium (t1 = 25 ºС).

Q2 (water heating) = С * m2 * Δt, where С – beats. heat capacity (assumed C = 4200 J / (K * kg)), m2 – mass (m2 = 250 g = 0.25 kg), Δt – change in temp. (Δt = 10 ºС).

С * m1 * (t – t1) = С * m2 * Δt.

500 * 0.2 * (t – 25) = 4200 * 0.25 * 25.

t = (4200 * 0.25 * 10 / (500 * 0.2)) + 25 = 130 ºС.