Newton's Law of Cooling When the Temperature of the Air Isn't Given

[Drakkith]

Homework Statement

You come across a dead body at 3 PM. Its temperature is 83.6 F. 30 minutes later its temperature is 78.6 F. How long was the guy dead before you found him?

Homework Equations

$\frac{dT}{dt}=-k(T-Tair)$
T(0) = 83.6 F
T(0.5) = 78.6 F

The Attempt at a Solution

This is a separable differential equation. Solving it gives me:
$T(t) = Ae^{-kt}+Tair$
However, I don't seem to be able to solve this completely. It feels like I haven't been given enough information (though I'm sure I have...).

At t=0:
$T(0)=83.6=A+Tair$
At t=0.5:
$T(0.5) = 78.6 = Ae^{-0.5k}+Tair$

Trying to solve for k by getting Tair in terms of A:
$78.6=Ae^{-0.5k}+83.6-A$
$Ae^{-0.5k}=A-5$
$e^{-0.5k}=1-\frac{5}{A}$
$-0.5k = ln|1-\frac{5}{A}|$
$k=-2ln|1-\frac{5}{A}|$
$k=ln(1-\frac{5}{A})^{-2}$

Using this in the 2nd equation where t=0.5:
$78.6=A(1-\frac{5}{A})^{-2}+83.6-A$
$A-5=\frac{A}{1-\frac{5}{A}}$
Dividing by A: $1-\frac{5}{A}=(1-\frac{5}{A})^{-2}$

The last equation is obviously not correct and I'm not sure what else to do. I've tried getting A in terms of Tair, but it doesn't seem to help. I've also tried using $\frac{dT}{dt}=-k(T-Tair)$ in various ways but I can't make anything work there either.

 

[PeroK]

I haven't looked at your equations, but one way to tackle the problem is to assume that the air is two different temperatures and see whether you get the same answer.

 

[DrClaude]

$k=ln(1-\frac{5}{A})^{-2}$

Using this in the 2nd equation where t=0.5:
$78.6=A(1-\frac{5}{A})^{-2}+83.6-A$

You made a mistake in going from that first equation to the second.

 

[Drakkith]

You made a mistake in going from that first equation to the second.

In fact, I made two. I left out the t=0.5 and I forgot a negative sign. I'll correct these and keep trying.

 

[Drakkith]

Also, it just occurred to me that I plugged k back into the equation I used to solve for it. That's not going to help. I'll just end up with 1=1.

 

[Mark44]

I'll just end up with 1=1.

True, but no help in this case...

 

[Drakkith]

True, but no help in this case...

Indeed.

 

[Drakkith]

Well, I haven't been able to solve this and I have other homework and studying that I need to move on to.
Thanks for helping gentlemen.

 

[PeroK]

Well, I haven't been able to solve this and I have other homework and studying that I need to move on to.
Thanks for helping gentlemen.

It seems fairly clear that you need the temperature of the air. If the air temperature was very low, then the cooling would be roughly equal before and after the body was found. But, if the air temperature was high, the cooling would have been much faster in the time before the body was found.

 

[Drakkith]

It seems fairly clear that you need the temperature of the air. If the air temperature was very low, then the cooling would be roughly equal before and after the body was found. But, if the air temperature was high, the cooling would have been much faster in the time before the body was found.

Indeed. It doesn't even tell me if the air temperature remained constant or not.
(Should I mark this solved??)

 

[PeroK]

Indeed. It doesn't even tell me if the air temperature remained constant or not.
(Should I mark this solved??)

It's up to you whether you think you can gain anything from doing the problem. The way I look at it, there's a general problem here about how to relate the initial time $t_1$ to everything else. (By the way, I would have set $t=0$ as the time of death.) Then, I would solve the general problem and just plug in the numbers for this specific case. If you don't have all the numbers (or any of the numbers) then it wouldn't matter to me.

In this case, we have $t=0$ is time of death, $t_1, t_2$ are the times of the two temperature measurements (we are trying to find $t_1$ and we know $\Delta t = t_2 - t_1$.

$T_0$ is the initial body temperature (above the air - this is something else I would do to simplify the algebra) and likewise $T_1, T_2$ are the two temperature measurents (relative to the air). That also translates to your differential equation: you have simply $T$, instead of the slightly clumsy $T-T_{air}$.

Then the probem is to show that:

$t_1 = \Delta t (\frac{ln(T_0/T_1)}{ln(T_1/T_2)}) \ \$ (1)

Then, you can sanity check this. E.g:

$t_1$ is related logarithmically to the body temperature (that looks good).

If $T_2$ is small (i.e. body has almost completely cooled at $t_2$), then the denominator is large and $t_1$ is small. That fits as well with negative exponential cooling.

At that point I'd probably be happy that I've got the right solution.

It's up to you if you want to try to derive equation (1).

 

[late347]

Indeed. It doesn't even tell me if the air temperature remained constant or not.
(Should I mark this solved??)

I think that it could be reasonable to assume ambient air remains constant temp, and body temp gets cooler.

We had a Newton's law of cooling assignment for logarithm class, but we didn't need calculus for that one!
I guess the formule that we had was simply the temperature differential, as a function of time.
d_0 was the temperature differential, at time stamp t=0
k was an unknown constant
and we had some values of temperatures for ambient and body temp
our problem was simply about using logarithm rules and getting the k first, and then finding out when the temperature of the body, becomes a certain temperature.
$d(t) = d_0e^{kt}$

so that was our formula we had for Newton's law of cooling.
so, if you have a heated house in the winter and inside it is +22C and outside it is -10C. The differential between those is going to be 32C. And this was at the t=0.
1/3 hours later, the inside temperature lowered by 1 C (but it was assumed outside it is still going to be -10C)
I think the idea is that ambient temperature won't be much affected by small transfer of heat into it.

 

[PeroK]

I think that it could be reasonable to assume ambient air remains constant temp, and body temp gets cooler.

I think the idea is that ambient temperature won't be much affected by small transfer of heat into it.

The air temperature might change over time for other reasons!

 

[late347]

The air temperature might change over time for other reasons!

yes, but weather patterns can indeed be bothersome for equation solving because of that reason exactly...

 

[Nidum]

Try a three point fit with T₀ = 98.6 F

(where T₀ is temperature at the actual time of death)