Chem. Find specific heat capacity.

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!

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  1. [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.

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  2. 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

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  3. You are right.

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