Question

A 2.61 g sample of a substance suspected of being pure gold is warmed to 72.8 ∘C and submerged into 15.8 g of water initially at 24.9 ∘C. The final temperature of the mixture is 26.1 ∘C. What is the specific heat capacity of the unknown substance (in J/g*ºC)?

Answer 1

mg = 2.42 g
Tg = 72.2 deg.C
mw = 15.8 g
Tw = 24.5
Tf = 27 deg. C
Cw = 4.18 J/g. deg. C

Energy balance for insulated system,
ΔE = 0
ΔUw + ΔUg = 0
ΔUw = - ΔUg
Qin = Qout
mg*cg*ΔTg = mw*cw*ΔTw
2.42*cg*(72.2 - 27) = 15.8*4.18*(27 - 24.5)
cg = 1.5095 J/g. deg.C or 1.5095 J/g.K