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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.
8 comments:
What is the rate of energy transfer by radiation from a metal cube 1.7 {\rm cm} on a side that is at 740^\circ {\rm{C}}? Its emissivity is 0.30.
i did, 5.67*10^-8 *0.3 * 0.017^2 *(270 + 273)^4...
what am i doing wrong??
i meant 740 plus 273 , sorry. i still dont get it though
Thats exactly what I was doing but i keep getting it wrong too. Can someone give a hint...
Okay we are doing everything right except for the area part. Since its a cube we have to keep in mind that it has 6 sides, and therefore we need to multiple the area by 6.
I hope this helps and you get it right =)
still not getting it with the given side multiplied by 6.
Why do we need to multiply the area by 6 when it specifically states that the cube is on its side? Doesn't that imply that only one side of the cube is relevant in transferring the heat by radiation? I'm confused..
You have to multiply it by 6 because that's the formula to calculate the Area of a cube (A=6a^2).Don't forget to convert the length given from cm to m.And also the Temp from C to K.
You also have to square A because it's area.
Power intensity I=(5.67e-8)*e*T^4
Then P=I*6A^2 (but make sure you convert A to m)
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