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Welcome to PHY 121 Blog Help. Here's how it works. For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
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12 comments:
Sorry, what do we use as the temperature of the burner?
That is what you need to find out
Oh right...sorry about that.
do we need the thermal conductivity for copper?
Yes
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
add that to the boiling temp of water- 100 degrees
OH MY GOD THANK YOU SO MUCH =)
To the above commentor:
I'm confused why you have to add it to 100 instead of subtracting it from 100. Enlighten me, please!
In his equation, T = Tf - Ti = .13 So Tf = T + Ti. .13 + 100. Ti = 100 because you start with boiling water in the kettle.
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!
I second that!
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