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 = mcÎ”t
mcÎ”t = mcÎ”t
(27.3 g)(clead)(26.3298.90)=(15.0g)(4.184)(26.3222.50)
clead = .121

Chemistry. Please check my answer!  DrBob222, Tuesday, December 2, 2008 at 8:40pm
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

Chemistry. Please check my answer!  Damon, Tuesday, December 2, 2008 at 8:42pm
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.

Chemistry. Please check my answer!  Sarah, Tuesday, December 2, 2008 at 8:50pm
Thank you both very much for your help and for responding as quickly as you did!

uhfjsb danq  uhfjsb danq, Saturday, January 17, 2009 at 3:51am
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