Two body applications of Newtons law of cooling.

In summary, 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 is T_a = [(T_a0 - T_b0)e^(-(1/C_a + 1/C_b)kt) + (C_aT_a + C_bT_b)/C_b]/(1 + C_a/C_b). This formula takes into account the change in temperature of the environment, unlike the formula T(t) = T_s + (T_0 - T_s)e^(-kt) which assumes the environment's temperature remains constant. It is derived using Newton's law of cooling and the definition of specific heat, and is
  • #1
SrEstroncio
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0

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 [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.


Homework Equations





The Attempt at a Solution



Thanks in advance for your help.
 
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  • #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.
 

Related to Two body applications of Newtons law of cooling.

1. What is Newton's law of cooling?

Newton's law of cooling is a principle that states that the rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings. This means that the hotter an object is compared to its surroundings, the faster it will cool down.

2. How is Newton's law of cooling applied to two bodies?

In two body applications of Newton's law of cooling, the principle is used to describe the rate of heat transfer between two bodies at different temperatures. This can be used to calculate the amount of time it will take for the bodies to reach the same temperature, or to determine the temperature difference between the bodies at a certain time.

3. What factors affect the rate of cooling in two body applications?

The rate of cooling in two body applications is affected by several factors, including the temperature difference between the bodies, the surface area of the bodies, and the type of material the bodies are made of. Other factors such as air flow and insulation can also impact the rate of cooling.

4. Can Newton's law of cooling be used to predict the temperature of two bodies over time?

Yes, Newton's law of cooling can be used to make predictions about the temperature of two bodies over time. By knowing the initial temperatures of the bodies and the rate of heat transfer, the law can be used to calculate the temperature of the bodies at any given time.

5. Are there any real-world applications of two body applications of Newton's law of cooling?

Yes, there are many real-world applications of two body applications of Newton's law of cooling. For example, this principle is used in food preservation to determine the optimal cooling rate for different types of food. It is also used in meteorology to predict the temperature of the Earth's atmosphere and in engineering to design cooling systems for electronics and machinery.

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