I just got a new AC rated at 6000 BTU and wanted to determine its power consumption. Some research on AC conventions quickly reveals that 6000 BTU really means 6000 BTU/h, where BTU is a measure of energy (British thermal unit). This is supposedly the rate at which heat is pumped from inside to outside. By the second law of thermodynamics, the power consumed to run the device must be even greater.

A direct conversion gives 6000 BTU/hr = 1758 W, so we should expect a power consumption greater than 1758 W. But I was surprised to see on the side panel of the unit (in the electricity specs) a voltage of 115 V and current of 4.66 A, suggesting a power of

$$115\text{ V} \times4.66\text{ A} \approx536\text{ W}.$$

At first I thought I must be misenterpreting the side panel, so I checked the official specs on the manufacturer's website. They boast a higher current of 6.5 A, and claim the electrical "rated input" of the device is 700 W. Still way too small. How is this possible?

P.S., all the sources I could find about this via googling seemed to contain similarly impossible numbers, without mentioning this seeming violation of the second law.


The industry rates air conditioner power consumption based on an average which takes into account the fact that once the air conditioner has got your room to the desired temperature, the compressor (which draws most of the power required to run the unit) is cycling on and off every few minutes, so the overall duty cycle or per cent time the compressor is actually running is typically around ~25%.

However, the heat transfer rate in BTU's assumes a 100% duty cycle, which is a confusing way of rating the performance of an air conditioner.

  • $\begingroup$ I don't think it's as simple as peak vs. average. I think, for electrical safety, the rated current on the label is the peak current. That's the value you'd use for a fuse or when deciding how thick the wires need to be. The real answer is more to do with the thermodynamics. $\endgroup$ – Oscar Bravo Jul 18 '19 at 6:54

Your given answer is correct. An air conditioner is a pump. The energy required to pump something is different from the energy required to create it afresh - think about, say, the energy required to pump water versus that to manufacture water from some kind of raw materials containing hydrogen and oxygen, or even by converting energy to mass(!). Clearly, the latter energies are typically considerably larger.

The difference here is that this pump is pumping energy, not mass. But the same principles apply - it does not need as much energy to move what it is moving as opposed to making it. However, since you cannot "make" energy, only get it from somewhere, the proper analogous quantity is the amount of energy moved. And the BTU ratings refer to this - how much energy is moved, as you say. See this.

To see if this scenario violates the second law of thermodynamics, we have to see what the second law of thermodynamics actually says regarding heat pumps, and compare this situation to that prediction, to see if there is a violation. To do this, we need a measure of the "efficiency" of a heat pump: how much energy is moved versus how much energy is expended to move it. This is called its coefficient of performance, or COP. For this unit, the COP is

$$\mathrm{COP} := \frac{\mbox{energy moved}}{\mbox{energy expended}} = \frac{1758\ \mathrm{W}}{536\ \mathrm{W}} \approx 3.28$$

The second law sets a maximum COP: the maximum amount of heat you can move between two reservoirs against a temperature gradient for any given amount of work. It is

$$\mathrm{COP}_\mathrm{max} := \frac{T_C}{T_H - T_C}$$

where $T_C$ and $T_H$ are the temperature of the cold and hot reservoir, respectively, which is called the Carnot COP, and is based on a "ideal" cooler which is called a Carnot cooler.

For the scenario of cooling down a room, $T_C$ will be the temperature of the room, and $T_H$ will be the temperature of the outdoors. Note that when the room temperature equals the outdoor temperature, the $\mathrm{COP}_\mathrm{max}$ is infinite - that is, an arbitrarily small amount of work suffices to create the initial temperature imbalance. However, as the temperature of the room drops (i.e. $T_C$ falls), the maximum COP drops as well. Hence, for a real AC with fixed COP, it will be unable to keep cooling once the maximum COP hits the AC's fixed COP.

Given a fixed COP and outdoor temperature $T_H$, one can find the minimum room temperature $T_C$ reachable satisfies, by equating $\mathrm{COP}_\mathrm{max} = \mathrm{COP}_\mathrm{given}$:

$$T_C = \frac{\mathrm{COP}_\mathrm{given}}{1 + \mathrm{COP}_\mathrm{given}} T_H$$

The absolute minimum temperature achievable with a COP of 3.28 and outdoor temperature of, say, $300\ \mathrm{K}$ is thus $230\ \mathrm{K}$, well below freezing. However, in practice, it will not be possible to reach this temperature because the ideal COP assumes the only point of heat flow is the machine itself, and perfect utilization of the energy input - at the calculated point, the heat going out from the machine will equal that coming back in through it due to the temperature gradient, a dynamic equilibrium. In reality, heat flow from the outdoors and the rest of the building through imperfectly-insulating walls, doors, and other portals will be adding additional heat back into the room. Moreover, a realistic machine will also waste input energy (e.g. electrical resistance in the wires, friction in the machine components, etc.), converting it into heat (not pumping heat) and hence acting to an extent as a heater for the room.

But the point is that, as long as the room temperature desired is above this temperature, then it is not a violation of the second law to cool to that desired temperature.


Never mind, I'm pretty sure my statement that the power required to operate an air conditioner (or any heat pump) must be greater than its rate of pumping heat is wrong. I.e., the second law doesn't forbid the coefficient of performance (heat pumped / energy consumed) from being greater than one.

  • $\begingroup$ I'm not sure you're wrong, I'd appreciate you keeping this open so we can see $\endgroup$ – TheEnvironmentalist Jul 18 '19 at 5:01
  • $\begingroup$ Second law only says the power consumed by your device can't be zero. That is, you'll need a finite amount of work to transfer heat from a cold to hot body. $\endgroup$ – Skawang Jul 18 '19 at 5:14

Clausius statement of the Second Law of Thermodynamics: Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.

Imagine a container of cold water which is placed inside room 1 in which temperature is higher than that of the water.
The temperature of the water will increase and in doing so the water will absorb heat from room 1 and so the temperature of room 1 will decrease.

Now do some work moving the container of water from room 1 to room 2 inside which the temperature is lower than that of room 1.
The temperature of the water will start to decrease and in doing so will heat room 2 and so the temperature of room 2 will increase.

So overall room 1 gets colder and room 2 gets hotter and for both heat transfers within the condensers it is from a hot body to a colder body.

A heat pump, your air conditioning unit, works on a continuous cycle as follows which is based on the Wikipedia article Heat pump.

enter image description here

  • 4 The working fluid, in its gaseous state, is pressurized and circulated through the system by a compressor.
  • 1 On the discharge side of the compressor, the now hot and highly pressurized vapour is cooled in a heat exchanger, called a condenser, until it condenses into a high pressure, moderate temperature liquid. Heat travels from condensor to the colder surroundings.
  • 2 The condensed refrigerant then passes through a pressure-lowering device also called a metering device. This may be an expansion valve, capillary tube, or possibly a work-extracting device such as a turbine.
  • 3 The low-pressure liquid refrigerant then enters another heat exchanger, the evaporator, in which the fluid absorbs heat and boils. Heat travels from the surroundings to the colder condenser. The refrigerant then returns to the compressor and the cycle is repeated.

The overall result is that some work is done by the compressor and heat is transferred from the surroundings in stage 3 to the surroundings in stage 1.

An advantage of such a system is that the energy expended compressing the working fluid is less than the amount of heat transferred so $1758 \,\rm J$ of heat can be "transported" (which you do not pay for) for the expenditure of $536\,\rm J$ of electrical energy (which you do pay for).

Another advantage is that the system can easily be reverse so instead of cooling a room it can be used to heat a room.


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