Estimating the heat capacity of my house

Friends, the spell of cold weather at the start of December 2022 has led to me breathlessly examining data on the thermal performance of the heat pump and the house.

During this period, outside temperatures fell as low as -5 °C and average daily temperatures were below 0 °C. In order to try to keep the internal temperature constant, I studied measurements of internal temperature taken every 2 minutes. The data were pretty stable, only rarely falling outside the bound of 19.5 °C ± 0.5 °C.

But looking in detail, I noticed a curious pattern.

Click on image for a larger version. Two graphs from the period 6th to 18th December 2022. The upper graph shows the air temperature in the middle of the house. At around 01:30 each night the temperature fell sharply. The lower graph shows the rate of change of the air temperature versus time (°C/hour). From this graph it is clear that the rate at which the temperature fell was approximately -0.95 °C/hour.

The upper graph shows sharp falls in temperature at 01:30 each night. These were caused by the heat pump switching to its hot water heating cycle. Prior to this, the heat flowing into the house from the heat pump was more-or-less balanced by the heat flowing out. But when the heat pump switches to heating the domestic hot water, there was no heating from the heat pump and the internal temperature fell.

The lower graph shows the rate of change of the air temperature (°C/hour) versus time over the same period. From this graph it is clear that the rate at which the air temperature fell during the domestic hot water cycles was approximately 0.95 °C/hour.

With a little mathematical analysis (which you can read here if you care) this cooling rate can be combined with knowledge of the heat transfer coefficient (which I estimated a couple of weeks ago) to give estimates of (a) the time constant for the house to cool and (b) the effective heat capacity of the house.

Analysis: Time Constant 

The time constant for the house, is the time for the temperature difference between the inside and outside of the house to fall to ~37% of its initial value after the heating is removed.

The time constant is estimated as (the initial temperature difference) divided by (the initial cooling rate). In this case  the initial temperature difference was typically ~20 °C and the initial cooling rate was 0.95 °C/hour, so the time constant of the house is roughly 21 hours. Sometimes it’s useful to express this in seconds: i.e. 21 x 3,600 = 75,600 seconds.

This suggests that if we switched off all the heating when the house was at 20 °C and the external temperature was 0 °C, the house would cool to roughly 7.4 °C after 21 hours. Intuitively this seems right, but for obvious reasons, I don’t want to actually do this experiment!

Note that this time constant is a characteristic of the house and does not vary with internal or external temperature.

Analysis: Thermal Resistance  

A couple of weeks ago I posted an analysis of the heating power required to heat our house as the ‘temperature demand’  increased as the external temperature fell. The summary graph is shown below.

Click on graph for a larger version. Graph of average heating power (in kW) versus temperature demand (°C) for the first 10 days of December 2022.

From this I concluded that Heat Transfer Coefficient (HTC) for the house was around 165 W/°C.

The inverse of the HTC is known as the thermal resistance that connects the inside of the house to the external environment. So the thermal resistance for the house is ~ 1/165 = 0.00606 °C/W.

Analysis: Heat Capacity  

A general feature of simple thermal analyses is that the time constant, thermal resistance and heat capacity are connected by the formula:

Time constant = Thermal resistance x Heat Capacity

Since we have estimates for the time constant (75,600 s) and the thermal resistance (0.00606 °C/W) we can this estimate the heat capacity of the house as 12,474,000 joules per °C.

This extremely large number is difficult to comprehend, but if we change to units more appropriate for building physics we can express the heat capacity as 3.5 kWh/°C. In other words, if the house were perfectly insulated, it would take 3.5 kWh of heat to raise its temperature by 1 °C.

We can check whether the number makes sense by noticing that the main mass of the house is the bricks from which it is built. A single brick weighs ~3 kg and has a heat capacity of ~2,400 J/°C. So thermally it looks like my house consists of 12,474,000/2,400 ~ 5,200 bricks.

However this estimate is too small. Even considering just the 133 square metres of external walls, if these have the equivalent of 120 bricks per square metre that would come to ~16,000 bricks.

So I think this heat capacity estimate just applies the heat capacity of the first internal parts of the house to cool. This refers to all the surfaces in contact with the air. So I think this is the effective heat capacity for cooling just a degree or two below ambient.

Why did I bother with this? The ‘Setback’ Problem

Friends, sometimes I go upstairs and forget why I went. And sometimes I start analysing things and can’t remember why I started! Fortunately, in this case, I had a really good reason for wanting to know the effective heat capacity of my house.

When I analysed the heat flows previously, I have had to assume that the temperature of the house was stable i.e. that there was a balance between the heat flowing in and the heat flowing out. As long as the temperature of the fabric of the house is stable, then it is neither storing or releasing heat.

However this isn’t enough if we want to understand some very common problems in the thermal physics of houses, such as “the setback problem”: this is the question of whether it’s smart to reduce the temperature of a dwelling (say) overnight and then to re-heat it once again in the morning. To answer this question we need to know about the rate at which a house cools down (it’s time constant) which is equivalent to knowing its heat capacity.

And that is why I have done this prolonged and tedious analysis. The next article will be an analysis of ‘The Setback Problem’. And it will be much more exciting!

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12 Responses to “Estimating the heat capacity of my house”

  1. Simon Duane Says:

    Interesting stuff, as ever – thanks Michael.
    I wasn’t quite sure – when you refer to “surfaces in contact with the air”, are you including the air itself? And, also, the rate at which air that flows through the house, carrying away heat by convection?
    (I don’t doubt that that is less significant for you in your house than it can be for me in my leaky house but, iirc, you do already have estimates of air changes per hour from your CO2 measurements a while back.)
    Best wishes

    • protonsforbreakfast Says:


      Good Evening.

      There are so many questions to consider!

      The heat capacity of all the air in the house amounts to around 0.1 kWh/°C so it’s relatively small compared to this ‘effective’ heat capacity.

      I think the air change rate will generally affect the heat transfer coefficient/thermal resistance.

      I think it must be losses of heat from the surfaces of all the walls/floors and objects in the house. It would be a great experiment to monitor that and if I have the energy I will do that!

      All the best


  2. Joe Wentworth Says:

    I look forward to the set back problem…
    I have been considering this and wondering about the cop variation between day and night air temprature. If the morning peak power after set back is acceptable with the better cop due to the warmer morning air? I will be interested to see what you find…

    I came across this and thought you might enjoy this level of data on the vaillant pumps in various conditions…

    • protonsforbreakfast Says:

      Joe, Good Evening.

      Thank you very much for the link to Vaillant Owner’s Page: all of a sudden I feel like I ‘belong’.

      Regarding ‘setback’, I have just published the article and as far as I see, the big difference concerns the relative efficiency of the heating in the steady state and in the ‘boost’ period that re-heats the house.

      In my model I had a single stable external temperature, but that could be modified. But the time when the ‘boost’ heating is required is generally early morning and that is generally the coldest part of the day – just before dawn.

      If the efficiency of ‘boost’ heating is less than the efficiency of ‘steady state’ heating then my calculations suggest that any energy savings will be small, or negative!

      I hope that helps


  3. Bob Pugh Says:

    As the outside coil (the evaporated) is effectively chilling outside air it needs to run a defrost cycle occasionally to keep the heat exchanger surface free of ice. I don’t know, but I would imagine this is done by bleeding hot gas from the condenser circuit into the evaporator, that’s how it would be done in industrial chillers. Some systems had electric heater elements built in to the evaporator coil to accomplish this, it was down to the individual manufacturers and designers.

    • protonsforbreakfast Says:

      Bob, Good Evening.

      As someone pointed out to me on Twitter, the defrost cycle is not all negative!

      When water vapour condenses as either water or frost on the condenser heat exchanger, it gives up its latent which can be considered as ‘bonus’ heating. But if the frost lingers and impedes heat flow, then the de-frost cycle is triggered. However this is really just ‘giving back’ the latent heat of condensation that had been previously captured!

      All the best


  4. David Edwards Says:

    ‘O’ Level Physics:
    12,474,000 joules per °C / 4200 Joules per Kg per °C = 2970.
    So your house has the same heat capacity as 3 tons of water.
    Is this helpful?

    • protonsforbreakfast Says:

      David, what a great example! That really makes me think.

      3 tonnes of water is 3 cubic metres of water.

      Mmmm. I am just trying to imagine this water spread in a thin layer over the inside of the house’ walls.

      If the layer were 0.5 cm think then the area would 600 m^2 – maybe that’s about right?

      I’ll reflect on this.

      Anyway, thank you for the comparison!


  5. Mark Purcell Says:

    Adding further complication to the “setback problem” consider calculating in terms of cost rather than energy.

    The cost of energy varies during the day based on time. Are there more benefits (savings) to run your heating hard during off-peak rates to enable you to run it low during peak rates. Insulation and thermal performance are going to skew calculations based on cost.

    • protonsforbreakfast Says:

      Mark, Good Morning,

      Ahh! Actual costs! I hadn’t thought about that.

      For heating with gas, I don’t think there are any timed tariffs, but for heating with electricity, I think costs generally don’t favour setback. Why? Because the normal times for setback are at night (when people are sleeping) and during the day (when people may be out of the home working or learning). But this means the ‘boost’ period is generally at times of peak demand (early morning) and (early evening) and costs are not generally low at this point.

      In any case, thanks for the thought!

      All the best


  6. Simon Says:

    To offer a more sophisticated comment…
    The house is a large and complicated system, and can be expected to have different dynamical behaviour, according to the time scale of interest. Your model, which is about as simple as can be, has just one timescale. You’ve adjusted your model so that it best represents the equilibrium behaviour of the house, and answers the question – what heating power is needed to maintain this fixed indoor temperature, given this unchanging outdoor temperature?
    You then compare what happens in the house, over a period of an hour or so, in response to turning the heating off, with what would happen in the model. Because the model has been set up using information about equilibrium, its dynamics will match the “slowest mode” of the house, whose timescale you’ve found to be many hours.
    If you did that experiment (perhaps you might both go away for a few days, in winter?) I think you’d find that the falling indoor temperature would be quite different from a pure exponential: lots of modes being relevant at first, and the slowest mode only dominating once many hours have elapsed.
    (a comment on the setback question may follow…)

    • protonsforbreakfast Says:

      Simon, Good Morning again!

      Yes, I agree the actual cooling curve would consist of many exponential decays. And I think by allowing the house to cool for only one hour I only observed the fastest of the decays. And that’s why the ‘effective’ heat capacity doesn’t match up with the real heat capacity of the house.

      All the best


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