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Two body applications of Newtons law of cooling.

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data

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

    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 [tex] T(t) = T_s + ( T_0 - T_s )e^{-kt} [/tex] 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.


    2. Relevant equations



    3. The attempt at a solution

    Thanks in advance for your help.
     
  2. jcsd
  3. Mar 10, 2012 #2
    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,

    [itex]\frac{dT_{a}}{dt}[/itex] = -ka(T[itex]_{a}[/itex]-T[itex]_{b}[/itex])

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

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

    [itex]\frac{dU_{a}}{dt}[/itex]=[itex]\frac{dU_{a}}{dT}[/itex]*[itex]\frac{dT_{a}}{dt}[/itex]=C*[itex]\frac{dT_{a}}{dt}[/itex],

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

    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;

    ∫[itex]\frac{dU_{a}}{\frac{U_{a}}{C_{a}}-\frac{U_{b}}{C_{b}}}[/itex]=∫-k*dt

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

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

    [itex]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[/itex]

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

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

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

    [itex] 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}}}[/itex]

    And sub in [itex]U_{a}=C_{a}T_{a}[/itex], so divide whole equation by Ca,


    [itex] 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}}}[/itex],

    With [itex]E=C_{a}T_{a}+C_{b}T_{b}=C_{a}T_{a0}+C_{b}T_{b0}[/itex], since E same at all times.

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

    So [itex]A=T_{a0}-T_{b0}[/itex] and we have our final form;



    [itex] 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}}}[/itex]


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

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

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

    And finally if both C's are equal, [itex]T_{a}=\frac{T_{a0}+T_{b0}}{2}[/itex]; 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.
     
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