Chem. Find specific heat capacity.
posted by Sarah on .
A piece of lead with a mass of 27.3 g was heated to 98.90oC and then dropped into 15,0 g of water at 22.50oC. The final temperature was 26.32oC. Calculate the specific heat capacity of lead from these data.
THANK YOU FOR ANY HELP!

[massPb x specific heat Pb x (TfinalTinitial)] + [massH2O x specific heat H2O x (TfinalTinitial)] = 0
Only one unknown; i.e., specific heat Pb. Solve for that. Post your work if you get stuck. 
This is what I did. Could you check my answer? I would really appreciate it.
qlead=qwater
q = mcÎ”t
mcÎ”t = mcÎ”t
(27.3 g)(clead)(26.3298.90)=(15.0g)(4.184)(26.3222.50)
clead = .121 
You are right.