In an iron vessel weighing 300 g, having a temperature of 10 * C, 200 g of water was poured at 20 * C

In an iron vessel weighing 300 g, having a temperature of 10 * C, 200 g of water was poured at 20 * C and a copper cylinder weighing 400 g was lowered at 25 * C. Determine the steady-state temperature.

m1 = 0.3 kg; t1 = 10 ºC; m2 = 0.2 kg, t2 = 20 ºC; m3 = 0.4 kg; t3 = 25 ºC; C1 = 460 J / (kg * K); C2 = 4200 J / (kg * K); C3 = 400 J / (kg * K).

1) C1 * m1 * (tр1 – t1) = C2 * m2 * (t2 – tр1).

460 * 0.3 * (tр1 – 10) = 4200 * 0.2 * (20 – tр1).

138tр1 – 1380 = 16800 – 840tр1.

978tр1 = 18180.

tр1 = 18.59 ºС.

2) (C1 * m1 + C2 * m2) * (tр2 – tр1) = C3 * m3 * (tр3 – tр2).

(460 * 0.3 + 4200 * 0.2) * (tр2 – 18.59) = 400 * 0.4 * (25 – tр2).

978tp2 – 18181.02 = 4000 – 160tp2.

1138tp2 = 22181.02.

tр2 = 19.49 ºС.