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A nichrome heating element across 230V supply consumes 1.5kW of power and heats up to 750*C. A tungsten bulb across the same supply operates at a much higher temperature of 1600*C in order to emit light. Does it mean that the tungsten bulb necessarily consumes greater Power?

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A typical incandescent light bulb with a tungsten filament consumes somewhere between $25$W and $100$W (much less than $1.5kW$) and reaches temperatures on the order of $2500^{\circ}C$ (much greater than $750^{\circ}C$).

So, why does it get hotter than a nichrome heating element?

It is because the temperature of a heated object depends not only on its power consumption, but also on its thermal resistance, which, in turn, depends on such factors as its surface area, air circulation, etc. The greater the thermal resistance of an object, the greater its temperature rise relative to the ambient temperature, given the same heat dissipation rate (which is close to the power consumption rate for the nichrome wire and somewhat lower for the tungsten filament, which radiates out some of the consumed power as light).

So, since, in this example, the power consumption of the tungsten bulb is substantially lower than the power consumption of the nichrome heating element, we have to conclude that the former must have much higher thermal resistance, e.g., because the diameter of the tungsten filament is smaller than the diameter of the nichrome wire.

In summary, the fact that a tungsten filament gets hotter than a nichrome heating element, does not mean that it consumes more power.

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  • $\begingroup$ I'm a bit surprised by the edit. The thermal conductivity is an intrisic property of materials, I do not see how it depends on the surface area. Could you precise a bit more this point? $\endgroup$ – thermomagnetic condensed boson Sep 21 '18 at 17:39
  • $\begingroup$ @coniferous_smellerULPBG-W8ZgjR Thanks for your comment: I agree. I've replaced thermal conductivity by thermal resistance, which does depend on the surface area. $\endgroup$ – V.F. Sep 21 '18 at 17:44
  • $\begingroup$ Only if you consider the heat transfer from the surface down to the inside of the wire, I think. Is that what you have in mind? $\endgroup$ – thermomagnetic condensed boson Sep 21 '18 at 19:36
  • $\begingroup$ @coniferous_smellerULPBG-W8ZgjR I am considering the thermal resistance of the whole path from the inside of the wire to the ambient air outside the bulb. It is a sum of a bunch of thermal resistances, including the internal resistance of the wire, which is negligible, resistance between the wire and the immediate gas atmosphere, which is significant and depends on the surface area of the wire, etc. $\endgroup$ – V.F. Sep 21 '18 at 20:30
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Power is the energy per unit time. When the object has reached an equilibrium temperature (which I assume are the temperatures you have stated), it is gaining heat energy (which is the power supplied by the source) at the same rate as it is losing it to the surroundings. This rate of heat loss is approximately proportional to the temperature difference, but it also depends on the geometry and physical properties of the object. So, in general, this does not necessarily mean that an object at a higher temperature requires more power to maintain its temperature. However, if they have similar geometries and materials, this should be true in most cases.

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No, it does not. Here is why:

Power is the product of voltage x current, and by ohm's law power is also equal to I^2 x R where I is the current and R is the resistance of the load.

What's different between your 2 examples is their resistance. The nichrome heater has a very low resistance compared to the bulb so it passes a lot more current and therefore dissipates much more power than the bulb: 1500 watts, as you state.

In comparison, a typical light bulb running on the same voltage dissipates 100 watts which means it draws much less current.

The operating temperature of each load does not enter into the calculation of the power.

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