A Derivation of Newton's Law of Cooling
and Implications for Water Heating
From the School of Hard Knocks. Last updated Nov 17, 2007
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Newton's Law of Cooling
Newton's Law of Cooling states that the hotter an object is, the faster it cools. More precisely, the rate of cooling is proportional to the temperature difference between an object and its surroundings. This word statement leads to the classic equation of exponential decline over time that applies to many phenomena in science and engineering, including the discharge of a capacitor and the decay in radioactivity.
Newton's Law of Cooling is useful when studying water heating because it can tell us how fast the hot water in pipes cools off, and also tells us how fast a water heater cools down if you turn off the breaker when you go on vacation.
The basic equation for Newton's Law of Cooling is:
T(t) = TA + (TH-TA) e-kt
Deriving the Equation:
Newton's Law of Cooling can be derived from two basic physics equations and a little calculus. Deriving the relationship from scratch is useful because we can then compute the coefficient k based on the R-value of the insulation and some other parameters, and compare that to the empirically measured value of k.
The first basic formula is that for heat loss through an insulating material:
H = A(Thot-Tcold)/R (1)
This formula can also be stated as:
H = AU(Thot-Tcold)
Here U is the conductance of the insulating material. The conductance is the reciprocal of the R-value. Both equations give the same result, they just have a slightly different form. Sometimes people who work with insulation find it more convenient to use R-values, and sometimes it is more convenient to use U values (depending on whether the insulation is in series or parallel).
I will use the first heat loss formula because we happen to have the R-values at hand.
The second basic formula relates the heat energy added to a material to its temperature:
Q = mc(T-Ti) (2)
Now imagine an insulated hot water tank containing a mass of hot water. Heat escapes through the insulation as described by the first equation, causing the water to cool down as described by the second equation. If we can link the two equations and apply a little calculus we can get some neat results. Here goes:
H = dQ/dt = A(Thot-Tcold)/R
Substitute T for Thot because that will be changing as the water cools, and substitute TA for Tcold because A stands for the ambient air temperature outside the insulation.
dQ = (A(T-TA)/R) dt
Restating equation (2) using differentials:
dQ = mc dT
We can now equate the two dQ's, but we have to be careful because one is stated as a positive and the other is really a negative, so we have to include a negative sign on one side:
mc dT = -(A(T-TA)/R) dt
dT/(T-TA) = -(A/(mcR)) dt
Taking the indefinite integral of both sides yields:
ln(T-TA) = -(A/(mcR))t + C
where C is the constant of integration.
Using each side as the exponent of e:
T-TA = C'e-(A/(mcR))t
where C' is actually a different constant: C' = eC
T(t) = TA + C'e-(A/(mcR))t
We can solve for C' by setting t=0:
T(0) = TA + C'
C' = T(0) - TA
But T(0) is actually the initial hot water temperature, call it TH
So C' = (TH - TA)
T(t) = TA + (TH - TA)e-(A/(mcR))t
Behold, this is Newton's Law of Cooling as stated at the top of this web page, with k=A/(mcR). We often know the values of A, m, c, and R, so can then compute k.
As a simple example, if I shut off the power to our 80 gallon R-16 water heater, how much would it cool down over the course of a day?
So the coefficient A/(mcR) = 37.5/(667*1*16) = 0.00351
Substituting into the equation:
T(t) = 60 + (120 - 60)e-0.00351 t
Setting t=24 hours:
T(24) = 115 F
A hot water tank with a higher R-value would cool down less, and one with a lower R-value would cool down more. As I stated on the web page on hot water tank heat loss, I suspect that the brass pressure relief valve and the hot water pipe may leak as much heat as the insulation. This means that the actual rate of cooling could be significantly faster than this calculated result.
This site is still under construction…to be continued…
You can e-mail me at support(@ sign goes here)leaningpinesoftware.com.