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Heat loss question

  • Thread starter EnrgyXprt
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  • #1
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How many watts would you need to keep a 64oz bottle of water at 40 degrees when the temperature surrounding the bottle is 10 degrees
(that's assuming you don't take into account surface area) ?


How many btus if you took into account the surface area of a 64oz soda bottle ?

What I did was,
1 BTU will raise the temp of 1 lb of water 1 Deg F

64 Oz bottle, 4 lbs of water

30 Deg Temp diff, 30 x 4 = 120 Btus
3.4 Btu's per watt

120/3.4 = 35.3 watts
 

Answers and Replies

  • #2
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Welcome to Physics Forums.

Is this the exact statement of the problem you are trying to solve? The statement makes no sense, particularly the part about not taking the surface area into account. You need to find the rate of heat transfer between the water inside the bottle, and the air outside. To do this you need to know the surface area, the heat transfer coefficient, and the temperature driving force (which you give as 30 F).

The way you did the problem is not correct. Also, 3.4 BTUs is not 1 watt. BTUs are energy, and watts are energy per unit time.
 
  • #3
rude man
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@ Chet: this has nothing to do with this post. I wanted to send you a WORD file within a personal message but can't do it, so I'm attaching the file here.

If you're interested, I have a writeup from a former co-worker who insists that a free expansion results in a temperature drop.

Any comment you'd like to make I will forward to him. I have given up trying to convince him he's wrong. Of course, if you agree with him, that'd be fine with me. I'm always willing to learn!

Thanks,
rudy
 

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