200 g of water is poured into a glass beaker weighing 100 g, the temperature of the water and the glass is 75

200 g of water is poured into a glass beaker weighing 100 g, the temperature of the water and the glass is 75, how much will the water temperature decrease when a silver spoon weighing 80 g is lowered into it at a temperature of 15.

mс = 100 g = 0.1 kg.

Cc = 840 J / kg * ° C.

mw = 200 g = 0.2 kg.

Cw = 4200 J / kg * ° C.

t1 = 75 ° C.

ml = 80 g = 0.08 kg.

Cl = 250 J / kg * ° C.

t2 = 15 ° C.

Δt -?

A certain amount of heat energy from water with a glass will be spent on heating the silver spoon. Therefore, the heat balance equation will have the form: Q = Ql, where Q is the amount of heat energy that the water with the glass will give while cooling, Ql is the amount of heat energy that goes to heating the silver spoon.

Q = Cc * mc * (t1 – t) + Cw * mw * (t1 – t).

Ql = Cl * ml * (t – t2).

Cl * ml * (t – t2) = Cc * mc * (t1 – t) + Cw * mw * (t1 – t).

Cl * ml * t – Cl * ml * t2 = Cc * mc * t1 – Cc * mc * t + Cw * mw * t1 – Cw * mw * t.

Cl * ml * t + Cc * mc * t + Cw * mw * t = Cc * mc * t1 + Cw * mw * t1 + Cl * ml * t2.

t = (Cc * mc * t1 + Cw * mw * t1 + Cl * ml * t2) / (Cl * ml + Cc * mc + Cw * mw).

t = (840 J / kg * ° C * 0.1 kg * 75 ° C + 4200 J / kg * ° C * 0.2 kg * 75 ° C + 250 J / kg * ° C * 0.08 kg * 15 ° C) / (250 J / kg * ° C * 0.08 kg + 840 J / kg * ° C * 0.1 kg + 4200 J / kg * ° C * 0.2 kg) = 73.8 ° C …

Δt = t1 – t.

Δt = 75 ° C – 73.8 ° C = 1.2 ° C.

Answer: the temperature of the water in the glass will drop by Δt = 1.2 ° C.