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