Answer = 35.3 Watts to Keep 64oz Bottle at 40 Deg in 10 Deg Temp

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SUMMARY

The discussion centers on calculating the wattage required to maintain a 64oz bottle of water at 40 degrees Fahrenheit in a surrounding temperature of 10 degrees Fahrenheit. The initial calculation suggests 35.3 watts based on a temperature difference of 30 degrees and the assumption of 120 BTUs needed for heating. However, it is clarified that this calculation is flawed as it neglects the heat transfer dynamics, including surface area and heat transfer coefficients, which are essential for accurate thermal analysis.

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  • Understanding of thermodynamics principles, particularly heat transfer.
  • Familiarity with BTUs and watts as units of energy and power.
  • Knowledge of heat transfer coefficients and their role in thermal calculations.
  • Basic physics concepts related to temperature differentials and energy transfer.
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  • Research the principles of heat transfer, focusing on conduction and convection.
  • Learn how to calculate heat transfer coefficients for different materials and conditions.
  • Explore the relationship between BTUs and watts in thermal energy calculations.
  • Study the impact of surface area on heat transfer rates in practical applications.
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Engineers, physicists, and anyone involved in thermal management or energy efficiency projects will benefit from this discussion, particularly those working with heating systems or thermal insulation.

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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
 
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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.
 
@ 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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