A Derivation of Newton's Law of Cooling

and Implications for Water Heating

From the School of Hard Knocks.  Last updated Nov 17, 2007

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) = T_A + (T_H-T_A) e^-kt

     where
       T(t) = Temperature at time t
       T_A = Ambient temperature (temp of surroundings)
       T_H = Temperature of hot object at time 0
       k = positive constant
       t = time
  
This kind of cooling data can be measured and plotted and the results can be used to compute the unkown parameter k. The parameter can sometimes also be derived mathematically.

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(T_hot-T_cold)/R         (1)

     where
       H = Heat loss in btu/hr
       A = Surface area in square feet
       T_hot = Hot water temperature in F
       T_cold = Air temperature in F
       R = R-value of the insulation in ft²hrF/btu

This formula can also be stated as:

     H = AU(T_hot-T_cold)

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-T_i)           (2)

     where
       Q = Heat in btus
       m = mass in pounds
       c = specific heat in btus/lb/F
       T = Temperature of the substance after the heat is added, in F
       T_i = Initial temperature of the substance in F

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(T_hot-T_cold)/R

Substitute T for T_hot because that will be changing as the water cools, and substitute T_A for T_cold because A stands for the ambient air temperature outside the insulation.

     dQ = (A(T-T_A)/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-T_A)/R) dt

Rearranging:

     dT/(T-T_A) = -(A/(mcR)) dt

Taking the indefinite integral of both sides yields:

     ln(T-T_A) = -(A/(mcR))t + C

where C is the constant of integration.

Using each side as the exponent of e:

     T-T_A = C'e^-(A/(mcR))t

where C' is actually a different constant: C' = e^C

So

     T(t) = T_A + C'e^-(A/(mcR))t

We can solve for C' by setting t=0:

     T(0) = T_A + C'

     C' = T(0) - T_A

But T(0) is actually the initial hot water temperature, call it T_H

So     C' = (T_H - T_A)

Finally:

     T(t) = T_A + (T_H - T_A)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.

Sample Calculation:

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?

Suppose:
       T_H = 120 F
       T_A = 60 F
       A = 37.5 ft² as computed in page on hot water tank heat loss
       m = 80 gal * 8.34 lb/gal = 667 lb
       c = 1 btu/lb/F for water
       R = 16

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.