Question

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

Answer 1

I obtained 0.12088 which rounds to 0.121 (your answer) if we have only three significant figures; that is, if the mass of Pb is 27.3 g and not 27.30 and the mass of the water is 15.0 g and not 15.00

Answer 2

Your method is correct.
I did not check the arithmetic but my physics book says lead is .130 so I think you did it right.

Answer 3

Thank you both very much for your help and for responding as quickly as you did!

Answer 4

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