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

- 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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