If I take three consecutive measurements $T_0$, $T_1$, $T_2$ at equal intervals, I can end up with two (Newton’s law of cooling) equations where the $e^{-kt}$ term is the same.

$$T_{n+1} = T_{outside} + (T_n - T_{outside}) × e^{-kt}.$$

So, without knowing the temperature constant of the building, can I calculate the outside temperature?

$$T_{outside} = \dfrac{T_0 × T_2 - T_1^2}{T_2 + T_0 - 2T_1}.$$

Have I missed something? Or is this really possible? (Provided no other sources of cooling and heating)

  • $\begingroup$ To clarify, to measure outside temperature indoor you mean to open the door for a consistent period of time, let the heat out, then close the door, wait for thermal equilibrium across the volume of the house, then measure and repeat? $\endgroup$
    – theUg
    Dec 11 '12 at 1:05
  • $\begingroup$ @theUg no, just measure temperature inside the house with no heating or cooling on. I'm assuming a less than perfectly insulated house. Or does that mean the newtonian equation is assuming perfect exchange? $\endgroup$
    – MandoMando
    Dec 11 '12 at 3:24

It's worth a try, but I can foresee two problems:

Firstly Newton's law of cooling is a good approximation for convective cooling of objects, but I don't know how well it would apply to a house. I'd have guessed most houses cool due to draughts and it's not clear whether this sort of cooling would obey Newton's law. Having said this, I would guess you still get an approximately exponential fall with time.

The more serious objection is that finding the equilibrium temperature by fitting an exponential decay gives very poor accuracy unless you have a large number of data points spanning a large temperature range. With only two points you're unlikely to get an accurate result.


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