1st Law of Thermodynamics in a Filament

For a filament lamp that is operating normally such that the temperature is constant, the rate of change in internal energy of the system is zero.

This would mean that rate of net heat transfer to the system + rate of work done on the system = 0

I can understand why rate of work done on the system will be positive since the electrons will do positive work on the ions in the filament as it collides with them. However, why is it that rate of heating of the filament is negative?

Wouldn't the rate of heating be zero? Since rate of thermal energy supplied by the source = rate of heat loss to surrounding by radiation

• The rate of heating of the filament isn't negative. You wrote it yourself: the rate of net heat transfer to the system is negative. The filament is heating its surroundings, corresponding to a negative number for transfer into the system. Nov 28 '21 at 5:20
• @Chemomechanics But isn't the rate of work done on the filament positive? So in order for change in internal energy to be zero wouldn't Q have to be negative?
– john
Nov 28 '21 at 5:24
• Q is negative. Q is the heating of the system, not by the system. Nov 28 '21 at 5:30
• Filament is not in thermal equilibrium with its environment, but rather in a steady state. So the laws of equilibrium thermodynamics do not apply, although one can still use the energy balance. Nov 29 '21 at 7:02
• Energy in equals energy out. Energy entering the system is electricity. Energy leaving is heat/light. Energy in we denote as "positive", energy out we denote as "negative". Zero energy flow, or equilibrium, happens when the positive flow of electrical energy balances the negative flow of heat energy.
– J...
Nov 29 '21 at 14:38

You have correctly deduced that the rate of heat transfer to the system is negative. This means that heat is leaving the system (that is leaving the filament).

What is heat? Heat is energy in transit from a higher temperature region to a lower temperature region due to the temperature difference.

In this case the filament is very hot and the surroundings are much cooler. There is a net flow of energy, largely in the form of infrared and visible electromagnetic radiation, from the filament to the surroundings. This is the heat flow from the system, that is heat leaving the system.

[Addendum Your confusion may be due to (understandable) confusion over words. I've given (above) the scientific meaning of 'heat'. You are using 'heating' to mean 'making hotter'. This is the everyday use of the word. And indeed, it's not silly, because usually supplying heat to something does make it hotter! But not always. You can supply heat to a pot of boiling water, but the water stays at 100 °C. And the electrically powered filament loses heat to its surroundings without getting cooler. In thermodynamics, avoid using 'heating' if you mean making hotter. You may use it to mean supplying heat, but it's safer to say 'supplying heat'!]

• Thanks for the answer! May I ask what would happen if the filament is in thermal equilibrium with the surroundings? In this case, there won't be a net flow of heat out or in to the system. So would that mean that internal energy will increase since Q=0 and work done ON the system is positive?
– john
Nov 29 '21 at 4:30
• @john yes, the internal energy (eg. temperature) of the filament would increase until it can loose enough energy once again to maintain a stable temperature OR until its internal energy is too much for the structure to handle (= the bulb burns out) Nov 29 '21 at 10:07
• Joined this site just to upvote your answer. Well-written! Nov 29 '21 at 17:23
• @ John I agree with Hobbamok, though the "thermal equilibrium with the surroundings" will be over as soon as you start to do electrical work, because, as we all agree, the filament's temperature will rise above that of its surroundings. Nov 29 '21 at 19:16
• @Nate Barbettini Thank you for your kind comment. I don't think you'll regret joining this site; it's full of good things! Nov 29 '21 at 19:18

Wouldn't the rate of heating be zero?

As @Chemomechanics very succinctly stated:

$$Q$$ is negative. $$Q$$ is the heating of the system, not by the system.

The 1st LoT may not be the most interesting description of what is going on in the filament, in terms of internal temperature profile e.g.

In this answer of mine (scroll down to below the break) I applied Fourier's Heat Equation for the steady state of a conductive cube with internal heat generation (aka 'load') and convective losses (note that use of symbols differs somewhat from yours) Although radiative losses maybe more appropriate for a hot filament (like a light bulb filament)

It would be very easy to apply it to a cylindrical geometry.

Many people have already answered the original question, and I made an account just now so I can't comment yet (I'll move this to a comment once I have the minimum reputation), but I'm responding to OP's follow-up question to Philip Wood's answer:

May I ask what would happen if the filament is in thermal equilibrium with the surroundings? In this case, there won't be a net flow of heat out or in to the system. So would that mean that internal energy will increase since $$Q=0$$ and work done ON the system is positive?

Yes, kind of, but to what extent depends on the relative heat capacities of the "surroundings" (which I'm going to call the "room" for illustration) versus "system" (filament). Call those heat capacities $$C_\text{room}$$ and $$C_\text{fil}$$.

If you've gotten the filament and room into a proper thermal equilibrium, of any additional work you do after this, a fraction $$\frac{C_\text{fil}}{C_\text{fil} + C_\text{room}}$$ goes into the filament. The rest goes into the room.

For what it's worth: to get to this point, you've utterly cooked the room!

(A caveat: I've assumed all internal energy changes in the filament + room are thermal. This is of course not strictly true—some of the energy has to go to light!—and there are plenty of microscopic weeds in which to get stuck. But the expression I gave is an upper bound on that fraction of additional work which goes into the filament's thermal energy. Given any "realistic" room and filament heat capacities, this is already an absurdly small number.)

• Hi thanks for the reply. Does this mean that theoretically, the filament can never be in thermal equilibrium with the surroundings? Because the moment it is in thermal equilibrium, Q=0 and the work done on the filament is positive so the internal energy will increase causing the filament to be in thermal disequilibrium again.
– john
Nov 29 '21 at 7:40
• This is a little hairier to answer—in particular, the $Q=0$ assertion isn't necessarily true. For simplicity say we start with both the filament and the room in thermal equilibrium, apply current to the filament for some time (imparting a total work $W$), and then stop and let the room and filament re-equilibrate. The filament presumably gets even hotter before it's allowed to equilibrate with the room; in that case, $\frac{Q_\text{fil}}{W} \leq \frac{C_\text{fil}}{C_\text{fil}+C_\text{room}} \neq 0$. In words: the filament's internal energy increases, but not by the entire amount $W$. Nov 29 '21 at 18:29
• But you could also say that after reaching thermal equilibrium, the filament and room maintain such efficient thermal contact that they're always in equilibrium (maybe the current is really tiny). Then while you might say $Q_\text{fil} = 0$, it's still not the case that all of the work done by the current goes into $\Delta U_\text{fil}$: instead, a (probably huge) fraction of it ultimately goes into the room instead. Here it's much more natural to think of the filament + room as the system, and write $\Delta U = W = \Delta U_\text{fil} + \Delta U_\text{room}$. Nov 29 '21 at 18:33
• Ultimately, if we allow the room and filament to equilibrate at all, we sort of lose the notion of a steady state temperature gradient between the filament and the room (which is actually an assumption you made in the original question). That relies instead on reaching a nonequilibrium steady state (NESS), as Roger Vadim pointed out—maybe you treat the room as an ideal heat bath with a temperature that doesn't change (this amounts to $C_\text{room} \to \infty$ in my original answer). In the ESS case, though, I think you'd be better served by taking the room and filament together as a system. Nov 29 '21 at 19:11
• In my first comment, I mistakenly wrote $\frac{Q_\text{fil}}{W} \leq \frac{C_\text{fil}}{C_\text{fil} + C_\text{room}}$. The expression should have a minus sign in front of it. Nov 29 '21 at 19:14

In case of a filament we are not dealing with an equilibrium state, but rather a non-equilibrium steady state - thus, in principle, one should not apply the laws of equilibrium thermodynamics to this case. However,

• to the extent that the 1st law of thermodynamics expresses the energy conservation, we can use the associated langauge to express the energy balance in the filament
• provided that the energy exchange between the filament and its environment is much slower than the relaxation processes within the filament, it can be treated as a system in thermodynamic quasi-equilibrium - i.e., we can describe it in terms of an effective temperature, etc.

The environment of the filament consists principally from the power source/circuit driving the current through it, and the space outside the filament, to which it emits photons. The power source does work on the filament at the rate given by the Joule-Lenz law: $$\frac{dW}{dt}=I^2R,$$ where $$I$$ is the current through the filament and $$R$$ is its resistance.

The filament then emits this energy as photons - the quasi-equilibrium assumption mentioned above allows us to treat the radiation emitted by the filement as thermal radiation and describe it by the black-body spectrum, which means that the rate at which the filament emits the radiation is given by Stefan-Boltzmann law: $$\frac{dQ}{dt} =-\alpha T^4,$$ where $$T$$ is the effective temperature of the filament, whereas $$\alpha$$ is the coefficient that depends on the material and the shape of the filament.

The rate of the internal energy change is then given by $$\frac{dU}{dt}=\frac{dQ}{dt} +\frac{dW}{dt}=-\alpha T^4+I^2R.$$ (One could also include heat conductance between the filament and its support, as well as the heat exchange between the filament and the surrounding gas, if it is not in vacuum.)

Initially, when the switch is turned on, the filament is a room temperature and the work done by current goes into heating teh filament, increasing its internal energy and temperature. As the temperature of the filament increases, it emits more and more radiation, until it reaches the steady state, where the energy lost via radiation equals to the work done by the current. Formally the steady state is given by condition $$\frac{dU}{dt}=0.$$ However we are not in a state of thermodynamic equilibrium, since there is constant energy flow from the power source to the radiation in the surrounding space. (One could use here a hydrodynamic analogy of water in a pond vs. a steady flow in a river.)

Remarks

• One could put the quasi-equilibrium description even a bit further by assuming that the internal energy is proportional to temperature (which is not granted in non-equilibrium) $$U=cT$$ and writing and solving the differential equation for the temperature of the filament: $$c\frac{dT}{dt}=-\alpha T^4+I^2R.$$
• One could also include the environment feedback. In typical conditions the radiation emitted into the environment is lost, e.g., via absorption by walls and other objects, which are in turn cooled via thermal conduction and convection. We could however imagine a lamp inside a thermally isolated box, which may be also assumed to be in quasi-equilibrium, characterized by temperature $$T_e$$. We can the write the rate equations for the energy of the filament and that of the box as $$c\frac{dT}{dt}=\alpha (T_b^4 - T^4)+I^2R,\\ c_b\frac{dT_b}{dt}=\alpha (T^4 - T_b^4).$$
• Thanks! your answer helped clear up some confusion I had. Can I clarify: does this mean that in a filament with current passing through, it is not possible for it to be in thermal equilibrium with the surroundings, but rather it can only stay at a steady state?
– john
Nov 29 '21 at 9:02
• @john This is correct. E.g., the filament temperature is clearly different from the room temperature - in equilibrium these two should be equal. Nov 29 '21 at 9:03

When you apply the first principle of thermodynamics, the definition of the system is important.

If we consider the system formed by the cylindrical conductor (ions and electrons) and the electromagnetic field inside, we are dealing with an open system in steady state. As the electrons enter and leave in the same state, there is no convective transfer associated with the movement of electrons and we can write that the internal energy variation of this open system is zero.

It remains to look for the energy transfers at the boundary of the system. What happens inside the system does not have to be taken into account and therefore, for this system, there is no internal heating to take into account. What remains is the flux of the Poynting vector entering the surface and the outgoing thermal flux, in the form of conducto-convective transfer and thermal radiation. The sum of these two terms is zero: the outgoing thermal power is the flux of the Poynting vector.

It is a classical exercise to show that the flux of the Poynting vector is equal to the Joule power $$RI^2$$