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Let us say you have an 8 watt source or heater, what is the highest temperature it can attain or be converted to?

I was asking in the context of transformer losses. At no-load, there is core loss.. if the temperature of the core is 50 Celsius. what minimum wattage of the iron core loss can produce that?

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    $\begingroup$ The problem is very-ill defined. We need (much) more information on your situation (what are you trying to heat, in what conditions, what is the environment, etc.........) $\endgroup$ – Barbaud Julien Nov 2 '18 at 1:57
  • $\begingroup$ It's the context of transformer no-load core loss.. as your know.. transformers heat up even with zero load at the secondary. So what is the loss of wattage.. or what minimum wattage can cause 50 Celsius surface temperature? $\endgroup$ – Samzun Nov 2 '18 at 2:12
  • $\begingroup$ Still far from being enough. If the losses happen in the core of the device, and you want us to compute an equilibrium temperature of the surface, we need at least to know the thickness of the material to be heated, its specific heat, the external conditions (outside ? what temperature ?), and probably the shape of your device if the convective transfer is to be accounted for. . Furthermore, I am not familiar at allwith the "no-load core loss" of a transformer, and maybe you'd have better luck in an electrical engineering stack exchange ? $\endgroup$ – Barbaud Julien Nov 2 '18 at 2:19
  • $\begingroup$ I'm not asking for the specific surface temperature or heat. I just want to know the physics principle of how to convert wattage to heat or temperature. If you have say a light bulb of 8 watts.. how much heat or temperature can it attain? $\endgroup$ – Samzun Nov 2 '18 at 2:21
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    $\begingroup$ I just want to make clear that your question has no answer. If your 8 watt light bulb is used to light up a dark alley in eastern russia during a warm winter evening at -40C, it definitely wont reach the same equilibrium as if it was in the bahamas. Again, your problem is not well defined, and a Power is completely unsufficient to answer. Furthermore, you are confusing the basic notions of heat and temperature, which will make the discussion delicate. Heat is an energy transfer. Temperature is not $\endgroup$ – Barbaud Julien Nov 2 '18 at 2:24
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The temperature will rise until it's dissipating heat at the rate heat energy is being put into it. If it were so well insulated that the temperature would have to be 20,000K for it to dissipate at that rate, then that is the temperature it would rise to!

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The temperature of an object is related to the total internal energy. And you've supplied one thing that will affect the total energy of this item: the fact that there is a power input of 8 watts (8 joules per second).

Unfortunately, this doesn't tell you anything about the other flows of energy (heat loss). If the object can easily lose that energy, then the temperature will remain very close to ambient. If well insulated, the temperature will need to rise a lot to force the energy out.

An analogy would be to learn that a constructed pool will be pumped with water at a rate of 8 cubic meters per second, and asked how high the water will reach. The answer depends on how the pool is constructed and how it holds water.

Likewise, the steady-state temperature of an object with a known power delivered to it depends on the object and its surroundings which control how the heat is removed.

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