Friday, November 26, 2010

Ch 13_3 #3

Use equation for the rate of conduction of heat across a temperature differenceand use for the temperature of the burner

13 comments:

Anonymous said...

Sorry, what do we use as the temperature of the burner?

Prof. Tsybychev said...

That is what you need to find out

Anonymous said...

Oh right...sorry about that.

Anonymous said...

do we need the thermal conductivity for copper?

Prof. Tsybychev said...

Yes

Anonymous said...

A copper-bottomed kettle, its bottom 24cm in diameter and 3.0 mm thick, sits on a burner. The kettle holds boiling water, and energy flows into the water from the kettle bottom at 800W.

Q/t=KA (T/L)
800W=(400W/mK)(pi .12m^2)(x/.003m)
800W=(400W/mK)(.045m^2)(x/.003m)
800=18(x/.003)
44=x/.003
x=.13 ?

Im stumped

Anonymous said...

add that to the boiling temp of water- 100 degrees

Anonymous said...

OH MY GOD THANK YOU SO MUCH =)

Anonymous said...

To the above commentor:

I'm confused why you have to add it to 100 instead of subtracting it from 100. Enlighten me, please!

Anonymous said...

In his equation, T = Tf - Ti = .13 So Tf = T + Ti. .13 + 100. Ti = 100 because you start with boiling water in the kettle.

Anonymous said...

5:52 PM Anon, thank you SO much. I was sitting here banging my head against what in the world I was doing wrong because I did all the calculations right!

Anonymous said...

I second that!

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