An iron piece weighing 300 grams, having a temperature of 10 degrees, and a copper plate weighing 400 grams

An iron piece weighing 300 grams, having a temperature of 10 degrees, and a copper plate weighing 400 grams at a temperature of 25 degrees are placed in 200 g of water at a temperature of 20 degrees. Find the steady-state temperature.

mw = 200 g = 0.2 kg.

tv = 20 ° C.

Sv = 4200 J / kg * ° C.

mw = 300 g = 0.3 kg.

tzh = 10 ° C.

Сж = 460 J / kg * ° C.

mm = 400 g = 0.4 kg.

tm = 25 ° C.

Cm = 400 J / kg * ° C.

t -?

Qzh = Qw + Qm.

Qw = Cw * mw * (tv – t).

Qzh = Czh * mzh * (t – tzh).

Qm = Cm * mm * (tm – t).

Cw * mw * (t – tzh) = Cw * mw * (tv – t) + Cm * mm * (tm – t).

Cw * mw * t – Cw * mw * tw = Cw * mw * tv – Cw * mw * t + Cm * mm * tm – Cm * mm * t.

Cw * mw * t + Cw * mw * t + Cm * mm * t = Cw * mw * tw + Cm * mm * tm + Cw * mw * tzh.

t = (Cw * mw * tv + Cm * mm * tm + Cg * mw * tg) / (Cw * mw + Cw * mw + Cm * mm).

t = (4200 J / kg * ° C * 0.2 kg * 20 ° C + 400 J / kg * ° C * 0.4 kg * 25 ° C + 460 J / kg * ° C * 0.3 kg * 10 ° C) / (460 J / kg * ° C * 0.3 kg + 4200 J / kg * ° C * 0.2 kg + 400 J / kg * ° C * 0.4 kg) = 19.5 ° C.

Answer: the temperature will be set at t = 19.5 ° C.