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Posted by on Sunday, November 30, 2008 at 8:25pm.

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!

  • Chem. Find specific heat capacity. - , Sunday, November 30, 2008 at 8:41pm

    [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.

  • Chem. Find specific heat capacity. - , Tuesday, December 2, 2008 at 12:31am

    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

  • Chem. Find specific heat capacity. - , Wednesday, April 30, 2014 at 9:25pm

    You are right.

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