Should I heat my room when I'm not here, energy-efficiently speaking?

I was wondering as it's getting cold : is it better for my electricity bill to shut down completely my (electric) heater during day, and to turn it on again when I come home (then it will have to heat the room from something like 5°C to around 20°C), or should I keep the temperature around, say, 15°C while I'm away ?

I would naively guess it's better off, but I'm also wondering about inertia of the walls for instance...

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This problem is very simple, but it's easy to overcomplicate by looking at too small a scale. At every second – no matter what the heater does – you waste money by heating the outside of your house. The rate of heating – and thus the rate at which you waste money – is given by Newton's Law of Cooling. So

$$\text{Wasted money} \propto \int (T_\text{in}-T_\text{out})\;dt$$

The lower your house's temperature, the less money you waste – no matter what. So set the thermostat to the lowest practical temperature when you're away.

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But there is also a requirement that it is habitable when you are in it. If it takes 6hours to reach 'normal' temperature when you get home in the evening that's not a solution –  Martin Beckett Feb 3 '12 at 18:40
Hence the "practical" bit I added. Basically, you'd want to lower the temperature a bit at first, then keep lowering it until you find a good trade-off between livability and savings. If you're comfortable shutting the heater off entirely, that's obviously the ideal – you don't need to measure your house's heat capacity in a series of silly experiments. –  rdhs Feb 3 '12 at 18:54
If the temperature drops below practical temperatures in a normal work day, you should really worry about the insulation of your house. –  Bernhard Feb 3 '12 at 19:24
There is another concern--besides how long it takes to come back up to habitable--in setting "practical": condensation is very bad. Make sure that doesn't happen. –  dmckee Feb 4 '12 at 3:11
Also in temperate climates during the day T_out may be high enough not to need much heating. So it would be a different comfort question if you are a night worker or come back at night. I know people who turn on the heat by phone two hours before going to their country cottage. –  anna v Feb 4 '12 at 9:23

This is a very easy question to answer, though as some have pointed out, it's also easy to get confused by making it too hard. If you think about it correctly, the answer is immediately apparent.

Let me ask you an exactly equivalent question:

Let us suppose I have a leaky bucket, one with a hole in the bottom. I'm going to leave for the day, and I'm trying to decide whether I waste less water by keeping the bucket full all the time I'm gone, or whether it makes more sense for me just to let the bucket run dry and then refill it when I get home.

The answer is obvious: I waste far more water trying to keep the bucket filled. I should turn the water to the bucket off until I get home, then refill it. Note that this is true for any amount of time I'm gone; I always come out ahead when I don't keep the bucket full, because the pressure differential from having a full bucket leads to the water exiting the bucket faster than if the bucket were only partially full.

Similarly, you should turn off the heat to your home while you're away (assuming there is no other reason to keep your house warm, e.g. a pet or cold-sensitive items), then reheat it when you get back. You will always save money this way, since as the heat differential between the inside of the house and the outside lessens, your heat loss will in turn lessen.

This is true in all but the most contrived cases, such as, "I have a well-insulated heat source in my house that will lose its insulative properties and dump all its heat if the temperature drops below X degrees". Ignoring such silly "gotcha" scenarios, the answer is good for all realistic cases.

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IMO it depends on the latitude. In northern regions and also up in mountains the pipes may freeze in the house and the expense of repairs will we worse than keeping the thermostat at a higher than freezing level. –  anna v Feb 4 '12 at 9:19
@annav I believe that your comment is already covered by the proviso: assuming there is no other reason to keep your house warm, e.g. a pet or cold-sensitive items. –  Paul Wagland Feb 7 '12 at 7:34

The answer to your question depends on a number of variables that are not so easy to estimate.

If you have relatively weak isolation in the walls and windows it is better to lower the temperature during the day quite a bit, how much depends also on how long you are gone.

If the heat capacity of the room and the things in the room is high it will take a long time in the evening to heat up again, so you don't want to cool it down too much.

If you have tight sealing windows and doors lowering the temperature significantly during the day will put your room temperature below the condensation point of the water vapor, so you will have wet surface on which mold will grow on quickly.

There are quite a few engineers working on these problems but there is no answer from the pure physics side what is better. It all depends on what you want to achieve.

To estimate an answer you can use an electricity meter connected to your heater and measure the room temperature over time when you switch the heater off, the power you need to keep the temperature constant and the heating rate and maximum power when you warm up your room again. With these values you can work out the heat capacity as $$C \propto {\tau}{\kappa}$$ with the relaxation time constant $\tau$ and the thermal conductivity $\kappa$ (details for example in this Calorimeter description).

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That's an interesting point you make about the dew point of vapor water, I didn't consider this side of the house-heating problem ! –  Zonko Feb 3 '12 at 14:10
The vapor water problem could be solved using an efficient MVHR (Mechanical Ventilation with Heat Recovery). See question physics.stackexchange.com/questions/23432/… –  Vincent Apr 9 '12 at 12:41

Let keep the problem as simple as possible. To start with some assumptions.

• The outside, or ambient temperature $T_a$ is constant.
• The temperature of your house $T_h$, is uniform over for your whole house, but a function of time $T_h(t)$. (Also known as an ideally stirred tank)
• The flow of heat out of your house is proportional can be cast into one heat transfer coefficient $h$ ($h$ will be smaller when your house is insulated better)

No heating The budget of heat for your house, contains two terms, the change in amount of heat in your house, and the flow out of your house.

$$c\frac{dT_h(t)}{dt}=-h(T_h(t)-T_a)$$ Where c is a constant related to the amount of heat that can be stored in your house (including the amount of heat stored in the walls, the size of your house, etc). This is easy to solve $T_h(t)=T_a+(T_h(0)-T_a)e^{-\frac{h}{c} t}$

Now, if you come back at your home, you want to heat your house from $T_h(t)$ back to $T_h(0)$. Your electrical heater should gives this amount of heat, which is just $c(T_h(t)-T_h(0))$, which equals to

$$Q_a = c \Big(T_a-T_h(0)\Big)\Big(1-e^{-\frac{h}{c}t_{away}}\Big)$$ where $Q$ is the amount of heat delivered by the heater and $t_{away}$ the time you've been absent.

Heater on When you keep your heater on the whole time. The amount of heat delivered by your heater, is constant and equal to $$Q_b=-h t_{away} (T_h(0)-T_a)$$

Now, if you are away for a short period of time, you can linearize the exponent, and you see that the amount of heat that you need to put in your house by the heater is the same to first order. However, as your away-time increases, the amount of electricity you need in the first case will reach a constant value, while if you canstantly heat your house the electricity you need will increase linearly.

The heat capacity of the walls you mention, will in principle affect the constants $c$ (related to total heat capacity) and $h$ (related to heat lose), will only change the timescale for which both approaches are more or less equivalent, but it will not change the physics or equations.

So, to conclude: you should determine the values of $c$ and $h$ of your house. You can measure h by trying the second method once. And then c by trying out the first method. Good luck!

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