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No-One's question at Yahoo! Answers regarding Newton's Law of Cooling

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MarkFL

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Feb 24, 2012
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Here is the question:

Newton's Law of Cooling extremely hard question?

Two bodies have been found. The wife was found dead inside the heated home where the temperature was maintained at 22 degrees Celsius. The husband dragged himself outside, where the outside temperature during the preceding day was 8 to 12 degrees Celsius.

The doctor took the temperatures of the body as soon as they arrived:
Wife: 33 degrees Celsius
Husband: 26 degrees Celsius

Who died first? Develop a model to help decide this question.

Help really appreciated.
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello No-One,

Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

\(\displaystyle \frac{dT}{dt}=-k(T-M)\) where \(\displaystyle T(0)=T_0,\,0<k\in\mathbb{R}\) and \(\displaystyle T>M\).

The ODE is separable and may be written:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrating, using the boundaries, and dummy variables of integration, we find:

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv\)

\(\displaystyle \ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt\)

\(\displaystyle kt=\ln\left(\frac{T_0-M}{T(t)-M} \right)\)

Assuming the heat transfer coefficient is the same for both bodies, which is a reasonable assumption since they are both human bodies, and that their body temperatures at the time of death was the normal 37° C., we have:

For the wife:

\(\displaystyle T_0=37,\,M=22, T(t)=33\)

\(\displaystyle kt=\ln\left(\frac{37-22}{33-22} \right)=\ln\left(\frac{15}{11} \right)\approx0.310154928303840\)

For the husband:

\(\displaystyle T_0=37,\,8\le M\le12, T(t)=26\)

\(\displaystyle \ln\left(\frac{37-8}{26-8} \right)\le kt\le\ln\left(\frac{37-12}{26-12} \right)\)

\(\displaystyle \ln\left(\frac{29}{18} \right)\le kt\le\ln\left(\frac{25}{14} \right)\)

\(\displaystyle 0.476924072090309\le kt\le0.579818495252942\)

Therefore, it is reasonable to assume the husband died first.
 

iamapineapple

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Aug 22, 2013
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The ODE is separable and may be written:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrating, using the boundaries, and dummy variables of integration, we find:

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv\)

\(\displaystyle \ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt\)

\(\displaystyle kt=\ln\left(\frac{T_0-M}{T(t)-M} \right)\)
Hey MarkFL, it's No-one from Yahoo! Thanks so much for answering my question in depth. It's just I've never used this method (above) for integration before. (i.e. multiplying by dt and then taking the integral?!). Would you mind explaining this step a little more indepth? Especially where you got the variables used in the definitie integral?

Much thanks already though. :)

- - - Updated - - -

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t \,dv\)
More just this step


So for the LHS

(I have no idea how to use latex for this)

\(\displaystyle \frac{1}{T - T_s} dT\)

Then put the integral sign?

\(\displaystyle \int\frac{1}{T - T_s} dT \)

\(\displaystyle Let u = T - T_s\)

\(\displaystyle \frac{du}{dT} = 1\)

so

\(\displaystyle \int\frac{1}{u - T_s} du\)

Is this correct? I don't get the RHS though

Thanks Mark

and why is the variable changed? (ha! sweet - killed the Latex stuff)

\(\displaystyle \int-k dt\)

\(\displaystyle k\int-1 dt = -kt\)
Correct?

Could you still please explain the limits though?
 
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MarkFL

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Feb 24, 2012
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Hello iamapineapple and welcome to MHB! (Sun)

Suppose you have the initial value problem (IVP):

\(\displaystyle \frac{dy}{dx}=f(x)\) where \(\displaystyle y\left(x_0\right)=y_0\)

Now, separating variables and using indefinite integrals, we may write:

\(\displaystyle \int\,dy=\int f(x)\,dx\)

And upon integrating, we find

\(\displaystyle y(x)=F(x)+C\) where \(\displaystyle \frac{d}{dx}\left(F(x) \right)=f(x)\)

Using the initial condition, we get

\(\displaystyle y\left(x_0 \right)=F\left(x_0 \right)+C\)

Solving for $C$ and using \(\displaystyle y\left(x_0\right)=y_0\), we obtain:

\(\displaystyle C=y_0-F\left(x_0 \right)\) thus:

\(\displaystyle y(x)=F(x)+y_0-F\left(x_0 \right)\)

which we may rewrite as:

\(\displaystyle y(x)-y_0=F(x)-F\left(x_0 \right)\)

Now, we may rewrite this, using the anti-derivative form of the fundamental theorem of calculus, as:

\(\displaystyle \int_{y_0}^{y(x)}\,dy=\int_{x_0}^{x}f(x)\,dx\)

Now, since the variable of integration gets integrated out, it is therefore considered a "dummy variable" and since it is considered good form not to use the same variable in the boundaries as we use for integration, we may switch these dummy variables and write:

\(\displaystyle \int_{y_0}^{y(x)}\,du=\int_{x_0}^{x}f(v)\,dv\)

This demonstrates that the two methods are equivalent.

Using the boundaries (the initial and final values) in the limits of integration eliminates the need to solve for the constant of integration, and I find it a more intuitive and cleaner approach to separable initial value problems.
 
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MarkFL

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Feb 24, 2012
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So for the LHS

(I have no idea how to use latex for this)

\(\displaystyle \frac{1}{T - T_s} dT\)

Then put the integral sign?

\(\displaystyle \int\frac{1}{T - T_s} dT \)

\(\displaystyle Let u = T - T_s\)

\(\displaystyle \frac{du}{dT} = 1\)

so

\(\displaystyle \int\frac{1}{u - T_s} du\)

Is this correct? I don't get the RHS though

Thanks Mark

and why is the variable changed? (ha! sweet - killed the Latex stuff)

\(\displaystyle \int-k dt\)

\(\displaystyle k\int-1 dt = -kt\)
Correct?

Could you still please explain the limits though?
Looks like you caught on rather quickly how to use $\LaTeX$! (Clapping)

If we did not use the boundaries in the limits, here is how we might solve the IVP:

\(\displaystyle \frac{dT}{dt}=-k(T-M)\) where \(\displaystyle T(0)=T_0\) and \(\displaystyle M<T\)

Separate variables:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrate:

\(\displaystyle \ln(T-M)=-kt+C\)

Using the initial values ($t=0$ and $T(0)=T_0$), we have:

\(\displaystyle \ln(T_0-M)=C\)

And so we find:

\(\displaystyle \ln(T-M)=-kt+\ln(T_0-M)\)

\(\displaystyle \ln(T-M)-\ln(T_0-M)=-kt\)

\(\displaystyle kt=\ln(T_0-M)-\ln(T-M)=\ln\left(\frac{T_0-M}{T-M} \right)\)

Since $T$ is a function of time $t$, we may finally write:

\(\displaystyle kt=\ln\left(\frac{T_0-M}{T(t)-M} \right)\)

And this is the same result we obtained via the other method. Whichever method you find more to your liking I suggest using, as you get the same result either way.

If anything I have posted is not clear, please do not hesitate to ask for clarification. :D
 

iamapineapple

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Aug 22, 2013
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I can't thank you enough, Mark. The second method was awesome, I'm a lot more familiar and comfortable with it, thanks so much!

I still have a few more question, sorry :)o)

So, I want to draw a conclusion from kt.
The smaller the value of k, the longer it would take for an object to cool (from my knowledge).

The wife's temperature (T(t)) is 33oC when the doctor checked her so dropped 4oC from her assumed body temperature of 37oC.

The husband on the other hand, dropped a whole 11OC

Based on the value of kt calculated, the husband had a faster rate of cooling than his wife: (wife, $kt = 0.31$ approx. and husband, $0.48\leq kt\leq 0.58$) so, the husband had a rate of cooling between $\frac{0.48}{0.31} = 1.55X$ and $\frac{0.58}{0.31} = 1.87X$ faster than his wife.

Therefore, the wife's temperature if she also crawled outside and was at her husband's cooling speed for as long as she was dead, would have had a temperature change between:

\(\displaystyle 4 \times 1.55 = 6.19 degrees\)
\(\displaystyle 4 \times 1.87 = 7.48 degrees\)

Therefore, because the husband's temperature dropped 11 degrees, greater than 6.19 and 7.48 degrees, it is reasonable to assume that the husband died first.

I'd appreciate your thoughts on this : )
 
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MarkFL

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Feb 24, 2012
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You are correct about the value of $k$. This constant of proportionality is called the heat transfer coefficient and is specific to the object (its composition and shape, specifically the ratio of surface area to volume) and independent of the ambient temperature $M$. Thus, this constant, along with the difference in temperature between the environment and the object, determines how quickly heat is transferred from the object to the environment.

When we apply Newton's Law of Cooling, we do so in cases where the ambient temperature is not significantly affected by the transfer of energy. For example, a human body, losing heat in the Earth's atmosphere, does not significantly change the temperature of the atmosphere. In the case of a body in a room, this change is controlled in this case by the thermostat attached to the heater keeping the room at a constant temperature.

If we were to put a cold spoon in a hot cup of coffee, we would expect that as the spoon heats up, the coffee also cools significantly, so we would want to account for that and use a more sophisticated model.

What you did in your comparison of the change in temperature between the husband and wife is valid, however, it suffices to merely compare the values of $kt$ for the two bodies given our assumption that $k$ is the same for both. Since the heat transfer coefficient is the same for both, we need only find that body for which the value of $kt$ is greater to say which had been dead for a longer period of time.
 
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MarkFL

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Feb 24, 2012
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There is another way to solve the ODE associated with the IVP, and that is to write it in standard linear form:

\(\displaystyle \frac{dT}{dt}+kT=kM\)

Compute the integrating factor:

\(\displaystyle \mu(t)=e^{k\int\,dt}=e^{kt}\)

Multiply through by this factor:

\(\displaystyle e^{kt}\frac{dT}{dt}+ke^{kt}T=kMe^{kt}\)

Observing that the left side is now the product of the derivative of the integrating factor and the dependent variable, we may now write:

\(\displaystyle \frac{d}{dt}\left(e^{kt}T \right)=kMe^{kt}\)

Integrate with respect to $t$:

\(\displaystyle e^{kt}T=Me^{kt}+C\)

Solve for $T(t)$:

\(\displaystyle T(t)=M+Ce^{-kt}\)

Use the initial conditions to determine the parameter $C$:

\(\displaystyle T(0)=M+C=T_0\,\therefore\,C=T_0-M\)

and we have:

\(\displaystyle T(t)=M+\left(T_0-M \right)e^{-kt}\)

Solving for $kt$, we find:

\(\displaystyle kt=\ln\left(\frac{T_0-M}{T(t)-M} \right)\)
 
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MarkFL

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Feb 24, 2012
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And here is yet another method we could use to solve the IVP (last one, I promise! (Tongueout))

Write the ODE in standard linear form:

\(\displaystyle \frac{dT}{dt}+kT=kM\)

Seeing the characteristic equation for the associated homogeneous equation is:

\(\displaystyle r=-k\)

We may then give the homogeneous solution as:

\(\displaystyle T_h(t)=c_1e^{-kt}\)

Next, observing that the differential operator defined by:

\(\displaystyle A\equiv D\)

annihilates the constant on the right side of the ODE, we may state:

\(\displaystyle D(D-k)[T]=0\)

And so we know the general solution will take the form:

\(\displaystyle T(t)=c_1e^{-kt}+c_2\)

Given the form of the homogeneous solution we already determined, we then know there must exist a particular solution of the form:

\(\displaystyle T_p(t)=c_2\)

Differentiating with respect to $t$, we have:

\(\displaystyle T_p'(t)=0\)

and then, using the method of undetermined coefficients, we substitute the particular solution into the ODE to get:

\(\displaystyle 0+kc_2=kM\,\therefore\,c_2=M\)

And so the particular solution is:

\(\displaystyle T_p(t)=M\)

And so, by superposition, we may state:

\(\displaystyle T(t)=T_h(t)+T_p(t)=c_1e^{-kt}+M\)

Determination of the value of the parameter $c_1$, and then solving for $kt$ is identical to the method used in my previous post.

The linear method I used in the previous post is quite useful when separation of variables is not possible for first order ODEs, and the method I used here in this post is useful for higher order ODEs.
 

iamapineapple

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Aug 22, 2013
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Thank you so much Mark, I can't express my gratitude enough.

Needless to say, I am extremely impressed by your level of knowledge. A request - would you mind helping me with questions in the future (if I have any)? My final exam is approaching and it'd be great to have someone like you to help.

Anyways, thanks so much.


-iamapineapple.
 

MarkFL

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Feb 24, 2012
13,775
Thank you so much Mark, I can't express my gratitude enough.

Needless to say, I am extremely impressed by your level of knowledge. A request - would you mind helping me with questions in the future (if I have any)? My final exam is approaching and it'd be great to have someone like you to help.

Anyways, thanks so much.


-iamapineapple.
I don't really have the time for one-on-one tutoring, but please feel free to post your questions here at MHB in the appropriate sub-forum (a question like this one would go in our Differential Equations sub-forum) along with your working so we know where you are stuck or can see what you have done to go astray, and I or one of our other helpers will be glad to offer assistance.

We have a great team of knowledgeable and friendly folks here who are glad to help.

Best Regards,

Mark.