Posted by Sarah on .
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,
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,
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,
Thank you both very much for your help and for responding as quickly as you did!

uhfjsb danq 
uhfjsb danq,
fwkyieqop nqoud lgdanpf tjcivs pdfokz bxvd npco