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Getting colder faster?

  1. Dec 8, 2003 #1
    This is very difficult to put into words clearly, so bear with me.

    You always hear that setting your thermostat higher expends more energy. My question is why?

    Let's assume we have two houses of the same size, type, surface area, and insulation (and in all other ways that i'm not thinking of the same).

    One house we set the thermostat at 78* and the other house we set at 68*. Also assuming a variance of 2*, such that the thermostat will drop to, say, 66* before it kicks on and warms the house to 70*, drops back down to 66* and so on.

    Just for kicks, let's say it's 50* outside(cuz it is right now)

    What extra energy(besides maintenance) would be used beyond the initial expenditure, what was required to heat the house up the extra 10*? The way I'm seeing it, (which is probably wrong) it shouldn't take any more energy after that point(besides maintenace, of course). Neither of the two houses should lose heat any faster than the other.
    That being the case the only 'waste' would be the additional energy required to bring the 1st house up to 78* in the first place. right?

    Am i missing any info here? or is the post clear and i'm not?
    Hit me back with anything i'm missing.
  2. jcsd
  3. Dec 8, 2003 #2


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    Heat loss is generally modelled by assuming that it is porpotional to the temperature difference.

    For your 2 equal homes, assuming the same outside temperature, the one which maintains a higher inside temperature will have a higher heat loss because of the higher temperature differential. This higher heat loss will mean that the furnace must run more to maintain the inside temperature.

    This of course corresponds to what you see in the real world, if you set your thermostat lower your heat bills also fall.
  4. Dec 8, 2003 #3
    Yeah I remember doing that stuff in maths last year, cant remember the exact formulae unfortunately but will try to find it when i get home. something to do with e to the power of t times the difference in temperatures or something, t being time. guess the easiest life situation you can put it into would be if you poured 2 cups of coffee into two cups exactly the same insulation etc. then you put milk in one so that it isnt as hot as the other. the temperature of the hotter one will drop quicker as the difference in temperature between the coffee and the external temperature is larger. whereas the temperature change of the cooler one will be much slower. correct me if im wrong - its been a while, but i think it is also an exponential curve. And the moral of the story is, if you want hotter coffee, put the milk in straight away not leave it for 5 mins and then put milk in. in relation to your question about energy, obviously if the hotter one cools down quicker, then its going to take more energy to keep it at the same temperature. DO AN EXPERIMENT!!
  5. Dec 9, 2003 #4
    This is it exactly.

    Here is the formula for Btu/hr loss through a wall.

    q=U x A x TD

    q = Btu/hr

    U = Heat transfer Coefficients (basically the reciprical of the R values) of the wall materials

    A = the wall area

    TD = the temperatue difference at base conditions.
  6. Dec 9, 2003 #5

    thanks guys

    i was laboring under the idea that all things equal(save for temperature) lose heat equally fast.
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