Asked by Sarah

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!
Answered by DrBob222
[massPb x specific heat Pb x (Tfinal-Tinitial)] + [massH2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Only one unknown; i.e., specific heat Pb. Solve for that. Post your work if you get stuck.
Answered by Sarah
This is what I did. Could you check my answer? I would really appreciate it.

-qlead=qwater
q = mcΔt

-mcΔt = mcΔt
-(27.3 g)(clead)(26.32-98.90)=(15.0g)(4.184)(26.32-22.50)
clead = .121
Answered by sally
You are right.
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