# Two body applications of Newtons law of cooling.

## Homework Statement

What's the formula that better describes the temperature as a function of time for an enclosed body of water with certain initial temperature $$T_a$$ immersed in another body of water of initial temperature $$T_b$$?

More clearly, I performed an experiment in which I put a small body of nearly boiling water (enclosed in a metal cylinder) inside a somewhat bigger body of ice-cold water, and measured T for both bodies of water at 1 minute intervals during 40 minutes; what's the equations that best describes this situation and to which I can compare my results? I am reluctant to use $$T(t) = T_s + ( T_0 - T_s )e^{-kt}$$ because the latter assumes that the environment's temperature remains more or less constant, while in my case I do measure a change of temperature (of about 10C) in it (the large body of cold water).

Sorry if my post was somewhat hard to read.

## The Attempt at a Solution

I wondered about this myself and decided to sit down and do it today out of curiosity. Post is 2 years old but it's worth putting. Also it's my first time using Latex so forgive the inconsistency of lettering.

Start from Newton's law of cooling, with t as time, T temp, and k constant,

$\frac{dT_{a}}{dt}$ = -ka(T$_{a}$-T$_{b}$)

Which we picked the sign of to make sure if T$_{a}$ >T$_{b}$, that T will decrease.

Now, we remember the definition of specific heat, C$_{a}$=$\frac{dU_{a}}{dT}$, then

$\frac{dU_{a}}{dt}$=$\frac{dU_{a}}{dT}$*$\frac{dT_{a}}{dt}$=C*$\frac{dT_{a}}{dt}$,

So our new Newton's law is $\frac{dU_{a}}{dt}$ = -$\frac{k_{a}}{C_{a}}$(T$_{a}$-T$_{b}$)= -k$(T_{a}-T_{b})$

And now we have to choose a T dependence for U. For Newton's law, we implicitly assumed U proportional to T, as is true for both the ideal gas and for metals from Dulong-Petit rule, so then we can express U=C*T, and separate variables for integration;

∫$\frac{dU_{a}}{\frac{U_{a}}{C_{a}}-\frac{U_{b}}{C_{b}}}$=∫-k*dt

$\frac{Ln[{\frac{U_{a}}{C_{a}}-\frac{U_{b}}{C_{b}}}]}{\frac{1}{C_{a}}-\frac{\frac{dU_{b}}{dU_{a}}}{C_{b}}}$=-k*t+c,

c integration constant for later. Now we employ energy conservation, so that
$U_{a}+U_{b}$=E, E constant. therefore 1+$\frac{dU_{b}}{dU_{a}}$=0, so our equation becomes;

$Ln[{\frac{U_{a}}{C_{a}}-\frac{U_{b}}{C_{b}}}]=(\frac{1}{C_{a}}+\frac{1}{C_{b}})(-k*t+c)=(\frac{1}{C_{a}}+\frac{1}{C_{b}})(-k*t)+d$

Exponentiate, and absorb the d as multiplicative integration constant. Add over $U_{b}$, multiply by $C_{a}$, and you get;

$U_{a}=C_{a}*A*e^{(\frac{1}{C_{a}}+\frac{1}{C_{b}})(-k*t)}+\frac{C_{a}}{C_{b}}U_{b}$

And to separate the coupled equations, we again use conservation of energy, so
$U_{b}=E-U_{a}$, add over the U term, factor out U, divide by coefficient, and;

$U_{a}=\frac{C_{a}*A*e^{(\frac{1}{C_{a}}+\frac{1}{C_{b}})(-k*t)}+\frac{C_{a}}{C_{b}}E}{1+\frac{C_{a}}{C_{b}}}$

And sub in $U_{a}=C_{a}T_{a}$, so divide whole equation by Ca,

$T_{a}=\frac{A*e^{(\frac{1}{C_{a}}+\frac{1}{C_{b}})(-k*t)}+\frac{E}{C_{b}}}{1+\frac{C_{a}}{C_{b}}}$,

With $E=C_{a}T_{a}+C_{b}T_{b}=C_{a}T_{a0}+C_{b}T_{b0}$, since E same at all times.

Then we apply initial conditions, at t=0 $T_{a}=T_{a0}=\frac{A+\frac{C_{a}T_{a0}+C_{b}T_{b0}}{C_{b}}}{1+\frac{C_{a}}{C_{b}}}$,

So $A=T_{a0}-T_{b0}$ and we have our final form;

$T_{a}=\frac{(T_{a0}-T_{b0})e^{(\frac{1}{C_{a}}+\frac{1}{C_{b}})(-k*t)}+\frac{C_{a}T_{a}+C_{b}T_{b}}{C_{b}}}{1+\frac{C_{a}}{C_{b}}}$

And we check some limits for sanity; at t→∞,
$T_{a}=\frac{C_{a}T_{a0}+C_{b}T_{b0}}{C_{a}+C_{b}}$,

So if $C_{b}$ goes to 0, ie you get no heat from changing it's temperature, then $T_{a}=T_{a0}$ ; it doesn't change temperature since there is no energy in the other system

If $C_{b}$ goes to ∞, it takes infinite energy to change it's temperature, and we see by L'Hopitals rule that $T_{a}=T_{b0}$; it raises to the unmovable temperature

And finally if both C's are equal, $T_{a}=\frac{T_{a0}+T_{b0}}{2}$; it reaches equilibrium at the average of the initial temperatures.

Nonlinear U dependences on T will need to be solved differently from the step listed as such, but for first order and most common systems this should work. Hope it helps.