Watts to Heat?

  • Thread starter burnit
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  • #1
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Main Question or Discussion Point

Hi All,

Could someone please help me with the calculations & an example of Watts consumed to Air Temperature rise in a given area:

Eg:
If we have a closed sealed container with no outside influences of area say 1 cubic meter, well insulated with an electrical heat source of the best efficiency supplying 100 Watts of heat inside.

What would be the temperature rise in Celcius for 1 minute?

Thank You
 

Answers and Replies

  • #2
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Sorry, i need to correct something, i should have said with 100 Watts being consumed from the Power supply & not with 100 Watts of heat in the container.
 
  • #3
russ_watters
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The specific heat of air is 716 J/kG K. The density of air is 1.3 kg/m^3. 1 watt is 1 J/sec.

So. In 60 seconds, you'd get a temperature rise of:
100*60/716/1.3=6.5 C
 
  • #4
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Thanks Russ,

So we assume 100% efficiency turning energy to heat with resistance heating elements etc.
 
  • #5
f95toli
Science Advisor
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No, we assume 100% inefficiency:wink:
Remember that heat is just "waste energy", when you use a certain amount of energy you always "produce" the same amount of heat.

In the case of a resistor there is nowhere else for the energy to go, so all of the energy you supple to the resistor is converted into heat.
 
  • #6
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Yes, i get what your saying with inefficiency.


I have another question with Watts Vs Heat in copper wire that i would like to be able to calculate & put some numbers to if someone could help me again with the equations & also an example.

Eg,

I have a length of stranded copper wire 300mm long consisting of 7 strands of "AWG 14" 1.63mm diameter each.
If i apply 100 Watts of power across this as in resistance heating, what will be the maximum surface Temperature of the wire if it stays connected to the 100 Watt supply or after 1 Min if time is applicable.
 
  • #7
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Not sure if AC or DC makes a difference but it is AC that i would like to calculate for.
 
  • #8
russ_watters
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The calculation is the same as what I just went through: just substitute the specific heat of copper in place of the specific heat of air.
 
  • #9
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Thanks Again Russ,

Just did a search & found an excellent chart of specific heat values, i wasn't sure how to go about this with different materials etc until now.

Learn something everyday on this Forum--Excellent!

Thank You Again
 
  • #10
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Just need to make sure i have this correct because i just did the calculation & it doesn't seem right?

Ok,

From the figures i found on the Net:

Specific Heat of Copper is 385 J/kg K, Density of Copper is 8940 kg/m^3

100*60/385/8940/m^3= .00174 deg C

Is this correct?
 
  • #11
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I Guess as in heating the air example it was heating a 1m^3 area, so the above calculation is also heating a 1m^3 block of Copper.
My wire has a smaller area, how do i calculate the area of my wire example into this equation?
 
  • #12
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After tossing & turning all night due to the lack of brain cells i'll give it another go & look at it in another way?

Heat required to raise the temperature of the example copper wire:

Mass of the example wire = .045 kg
Specific Heat Capacity of Copper = .39 KJ/kg K
Final Temp = 100 C
Initial Temp = 40 C

Q = mc(T2 - T1)

Q = .045 * .39 * 60 = 1.053 KJ/sec = 1053 Watts

Is this correct?
 

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