Question

A pure gold ring and pure silver ring have a total mass of 16.2 g . The two rings are heated to 78.9 ∘C and dropped into a 15.0 mL of water at 21.3 ∘C. When equilibrium is reached, the temperature of the water is 23.8 ∘C.

What is the mass of gold ring? (Assume a density of 0.998 g/mL for water.)

What is the mass of silver ring? (Assume a density of 0.998 g/mL for water.)

Gosh, I cant get the right answer here, please help (I am getting 4.6 g or 30 g, I am not sure)

Answer 1

Qgold = mg*Cg*(Tf-Tg)

Qsilver = ms*Cs*(Tf-Ts)

Ts = Tg = 78.9

Tw = 21.3

Tf = 23.8

Qwater = m*C*](Tf-Tw)

Qfrom metals:

Qmetals = mg*Cg*(Tf-Tg) + ms*Cs*(Tf-Ts)

Qmetlas = (mg*Cg + ms*Cs)(Tf-Ts)

Qmetals = (mg*0.129+ ms*0.24)(23.8-78.9)

Qwater = 15*4.184*(23.8-21.3)

Qmetals = -Qwater

(mg*0.129+ ms*0.24)(23.8-78.9) = 15*4.184*(23.8-21.3)

(mg*0.129+ ms*0.24) = ( 15*4.184*(23.8-21.3)) /(23.8-78.9)

(mg*0.129+ ms*0.24) = 2.84754

Mtotal = 16.2

mg = 16.2 - ms

(mg*0.129+ ms*0.24) = 2.84754

(16.2 - ms)*0.129+ ms*0.24 = 2.84754

0.129*16.2 + (0.24-0.129)ms = 2.84754

ms = (2.84754-0.129*16.2 )/(0.24-0.129) = 6.82648 g of silver

so

mass of gold = 16.2-6.82648 = 9.37352 g