posted by Sue on .
A 355-g piece of metal at 48.0 degree celsius is dropped into 111g of water at 15.0 degree celsius.If the final temperature of the mixture is 32.5 degree celsius what is the specific heat of the metal?
c1•m1 •Δt1 = c2•m2• Δt2
c1 = c2•m2• Δt2/ m1 •Δt1 =
=4180•0.111•17.5/0.335•15.5 = 1564 J/kg•oC