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The question in the title is of course childish and the answer is NO...
But then how much if not 50 degree ?

lets give some more data:
we have a electric heater with NO temperature regulation and only one stage On or Off.
It is used to heat a small room and if the outside temperature is 0, the room temperature stays around 25.

How will the room temperature change if we add another electric heater of the same type ?

d.raev
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  • You don't know, because you don't know the thermal dissipation to the outside. It might be linear with $ \Delta T $ or it might not. If it is, then if also assuming the air in the room has a constant temperature vs. energy content, it's not too hard. – Carl Witthoft Jan 29 '14 at 15:48

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If the outside temperature – on all sides of the room – is 0 degrees and one heater (which is much hotter than 50 degrees) will be able to reach 25 degrees in the room, two heaters will reach 50 degrees. That's true for the temperature change over a particular timeframe; it is also true for the equilibrium temperature (the temperature we reach after some time when the leaking heat cancels the heat flowing from the heating device).

The reason is that all the heat fluxes are (almost exactly) proportional to the temperature differences and the heat fluxes are equal to the extra heat coming from the heating device which is also doubled.

In reality, the final or equilibrium temperature will be lower than 50 degrees because of some of the reasons below or their combination:

  1. A part of the warmth in the room isn't due to the heating system but due to surrounding apartments or other spaces whose temperature is higher than 0 degrees. One may say that the average outside temperature is higher than 0 degrees. If it is $T_a$, then two heating devices will achieve $T_a+2\times(25-T_a) = 50-T_a$ which will be lower than 50 degrees if $T_a$ is greater than zero. Alternatively, we may say that the heating is just a part of the temperature increase from 0 degrees, so doubling the heating makes a smaller impact.

  2. The heating itself isn't much warmer than 50 degrees. For example, if the heating's own temperature is 40 degrees, then 40 degrees is also the maximum temperature one may achieve in the room even if we combine many heating devices of the same kind. The temperature will only converge to 40 degrees from below, so the addition of the temperature differences will be sublinear.

  3. When it gets much warmer than 25 degrees, someone will find it really annoying and will try to turn the heating off.

  4. Some insulators around the room may have heat conductivity that (usually slighty) increases with the temperature, so the heat losses at higher temperature will be greater than expected from the proportionality to the temperature difference.

Luboš Motl
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  • I disagree on several counts. First, just stick with the simple problem: a standalone room. Second, the "heater" is an energy source. It doesn't matter what its local temperature is: it's just dumping X joules/second into the room. – Carl Witthoft Jan 29 '14 at 15:54
  • Carl, sorry, it doesn't. There is no device that can do what you say. The reason is called the second law of thermodynamics. If the room becomes warmer than heater, there can't be any transfer of heat from the heater to the room: heat always goes to the cooler body. In fact, the flow will go in the opposite direction. This excludes your rule for the usual warm-water heaters. Other types of heaters, including direct electric ones, will also show a decrease of transferred energy at higher temperature, although not as dramatic as the usual warm-water heaters. – Luboš Motl Jan 29 '14 at 15:58
  • Thanks for the detailed answer and I appreciate the extra notes. Can you add the formula/relation for this problem? I thought it wont be linear like this... – d.raev Jan 29 '14 at 16:03
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    Dear D. Raev, well, the heat transfer is almost exactly equivalent to the temperature difference (search for "temperature difference" at https://en.wikipedia.org/wiki/Heat_transfer ) which means $dE/dt = C\cdot (T_{\rm out}-T_{\rm in})$ with some constant $C$ which depends on the setup, materials etc. in a complex way. The linearity in $\Delta T$ is inherited from heat conduction (diffusion) equation. – Luboš Motl Jan 29 '14 at 16:09
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    Everywhere here, it is also assumed, and it is approximately right, that the change of the thermal energy is proportional to the change of the temperature, $\Delta E = C \cdot\Delta T$, where $C$ is the heat capacity. This is equivalent to the assumption that the heat capacity doesn't depend on the temperature (too much) over the interval of temperatures that are probed by your process. For your "doubled heater" problem, one doesn't have to be told the values of the coefficients $C$ etc. because they cancel. – Luboš Motl Jan 29 '14 at 16:10
  • If you've got an electric motor being driven by an external source, please explain how energy can be pumped back into the motor regardless of temperature. If the "electric heater" heats the room to its current temperature but power is still being dissipated in the resistive element, everything including the heater will get hotter. There is no a priori reason to believe there's a maximum temperature the heater can reach. – Carl Witthoft Jan 29 '14 at 18:02
  • Well, above some temperature, the motor either breaks or melts down. ;-) But even before that, a direct electric heater with a resistor $R$ has output $U^2/R$ and because the resistance increases with temperature, https://www.google.cz/search?q=resistance+function+temperature&um=1&ie=UTF-8&hl=en&tbm=isch&source=og&sa=N&tab=wi&biw=1317&bih=708 , the power will go down as the temperature goes up. I didn't claim that every heater has a limiting temperature. Heating by warm water (which I have at home) surely has - the water has some temperature. The decrease of heating with temperature is univ. – Luboš Motl Jan 29 '14 at 18:55