Question

A pure gold ring and a pure silver ring have a total mass of 14.9 g. The two rings are heated to 62.0 ○C and dropped into 15 ml of water at 23.5 ○C. When equilibrium is reached, the temperature of the water is 25.0 ○C. What is the mass of each ring?

Answer 1

mg + ms = 14.9

T1 = 62

V= 15 ml

m = 15 g of waqter

Tw = 23.5

Tf = 25

then

Qwater = m*cp*(Tf-Ti) = 15*4.184*(25-23.5) = 94.14 J

then

Cp silver = 0.240

Cp gold = 0.129

Qgold + Qsilver = -Qwater

mg*Cpg*(Tf-Ti) + ms*Cps*(Tf-Ti) = -94.14

mg*0.129*(25-62) + ms*0.240*(25-62) =  -94.14

(0.129mg + 0.24ms) (-37) =  -94.14

(0.129mg + 0.24ms) = 2.54432

we have 2 euqtions

mg + ms = 14.9

0.129mg + 0.24ms = 2.54432

solve for ms

ms = 14.9 - mg

substitute

0.129mg + 0.24( 14.9 - mg) = 2.54432

0.129mg +3.576 - 0.24mg =  2.54432

(0.24-0.129)mg = 3.576 -2.54432

mg = (3.576 -2.54432)/((0.24-0.129)) = 9.294414

then

ms = 14.9-9.294414 = 5.605586