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December 20, 2014

Homework Help: Chemistry. Please check my answer!

Posted by Sarah on Tuesday, December 2, 2008 at 8:34pm.

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.

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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