Question

The temperature of a dead body that has been cooling in a room set at 70F is measured at 88F. One hour later, the body temperature is 87.5F. How long before the first measurement was the time of death, assuming the body temperature of the deceased at the time of death was 98.6F?

Answer 1

This is a *morbid* question, lol. I'm not sure if your teacher wrote this question themselves, but that's funny in a macabre kind of way.

Is this a differential equations question, do you have to use something like Newton's Cooling Law, or are you given any additional information at all?

Answer 2

@Mendicant_Bias Yes, it is a Newton's Cooling Law

Answer 3

How much is it decreasing an hour?

Answer 4

@sammixboo sorry, it didn't say it

Answer 5

No, it says it was 88 one hour, and decreased down to 87.5 after an hour, so how much did the temp decrease in an hour?

Answer 6

^That's now how you solve this problem, if I think you're gonna try to do it without Newton's Cooling Law. @ganeshie8 , could you help us out on this? I'm not familiar with applications of DE's, but I'm going to open up my book and take another shot at it.

Answer 7

*not how you solve this problem

Answer 8

@sammixboo : its not hour its 88 degree fahrenheit

Answer 9

Yes I know it is 88 degrees, :)

Anyways you seem kinda confused with what I am explaining, but this can help you http://formulas.tutorvista.com/physics/newton-s-law-of-cooling-formula.html

Answer 10

@sammixboo this is too confusing

Answer 11

Well, first thing you want to do is write out all your givens, right? If you're using `Newton's Law of Cooling`,It would be best to match your variables with the equation at hand. No?

Atleast that's the first step to overcoming confusion :)

Answer 12

@Jhannybean yea, the newton cooling law is confusing i still dont know which formula to use

Answer 13

I keep getting offtracked. You mind me asking what course this is for, specifically? Is it ODE, or is it a Physics class?

Answer 14

@Mendicant_Bias Its for college algebra...kinda difficult for me

Answer 15

Oh, okay, this can't be too bad, if it's just a given plug and solve sort of deal.

Alright, I think I know how to do this now. Give me one moment.

Answer 16

\[\frac{dT}{dt}=k(T_{t}-T_{s})\]

I think you can calculate dT/dt from your givewn values.

Answer 17

Man, the way this prompt is worded is just terrible. Sorry that I'm slow on this, it just isn't very easy to figure out based on how they are talking about this.

"The temperature of a dead body that has been cooling in a room set at 70F is measured at 88F. One hour later, the body temperature is 87.5F. How long before the first measurement was the time of death, assuming the body temperature of the deceased at the time of death was 98.6F?"

Alright, picking apart this prompt to find what values you are given and thus figure out how to solve it:




""The temperature of a dead body that has been cooling in a room set at 70F...". This tells us that our surrounding temperature is \[T_{s}=70^{\circ} \rm F.\]
"...assuming the body temperature of the deceased at the time of death was 98.6F?" This tells us that our initial temperature was\[T_{0}=98.6^{\circ} \rm F\]

Answer 18

Figuring out the remaining values momentarily. It could also just be that I'm bad at reading word problems, lol. I think we first need to use the other values, the ~80 ones, to solve for that constant, k, on second thought.
\[\frac{dT}{dt}=k(T_{t}-T_{s})\]
\[\frac{dT}{dt}=\frac{\triangle T}{\triangle t}=\rm \frac{87.5 \ F -88.0 \ F}{3600 \ s-0 \ s}\]

Answer 19

Does that last part make sense? I'm not entirely sure I'm allowed to do that and am just generally having trouble reading that prompt, but there's no harm in trying and making mistakes; better than doing nothing.

Answer 20

But yeah, does that last part make sense to you?

Answer 21

@Mendicant_Bias Yes! you really saved me xD Thank you so much!
but whats that triangle

Answer 22

Alright! I figured out how you are supposed to approach it, I was wrong initialll. Take a look at this, and let me know if this makes sense with the third equation in that link the othert user posted. \[87.5=70.0+(88.0-70.0)e^{-k(3600)}\]

Answer 23

What you're supposed to do is use the information you are given to solve for that rate constant, k, before anything else. The thing that is confusing about this problem is that they give you three different temperature values, so you might not know which to use, but only two have the proper information you need to solve for one variable. I'll make what I'm saying a little more clear in a second.

Answer 24

good! very good

Answer 25

@Mendicant_Bias yes this makes a lot more sense! thank you so much :D you saved me from difficulty lol

Answer 26

Let's say I tried to solve straight for the time between the first measurement and the second measurement, which is the final stretch of our problem. Taking that equation for Newton's Cooling Law, \[T(t)=T_{s}+(T_{0}-T_{s})e^{-kt}\]Where \[T(t)=\text{Final temperature},\]\[T_{0}=\text{Initial temperature},\]\[T_{s}=\text{Temperature of surroundings},\]\[k=\text{Rate constant, and}\]\[t=\rm time;\]
We would plug in values like such: \[88 \ \text{F}=70 \ \text{F}+(98.6 \ \text{F}-70.0 \ \text{F})e^{-kt}\]

The problem with this is that you have two unknowns! There's no way you can solve this with the given information. So, you have to solve for k using the other set of temperatures, where T(t)=87.5, and T_0 = 88.0.

Answer 27

Let's set it up now for the second set of values, one moment.

Answer 28

\[87.5 =70+(88-70)e^{-k(3600)}\]

Answer 29

Now you just need to solve for k. Can you take a shot at that?

Answer 30

@Mendicant_Bias would the answer be k=7.825?

Answer 31

\[87.5=70+(88-70)e^{-k(3600)}\]\[17.5=(88-70)e^{-k(3600)}\]

Exponentiate both sides.\[\ln(17.5)=\ln(18\cdot e^{-k(3600)})=\ln(18)+\ln(e^{-k(3600)})\]
\[\ln(17.5)-\ln(18)=\ln(e^{-k(3600)})\]\[-0.0281...=\ln(e^{-k(3600)})=-3600k\]

Answer 32

I feel like I have to have made a mistake somewhere, let me double check my own work really quickly.

Answer 33

\[0.97222...=e^{-k(3600)}\]
\[-0.0281...=-3600k\]

Nah, I did it right. Your k value should be: \[\frac{-0.0281}{-3600}=7.82\cdot10^{-6}\]

Let me know if you want me to work through the algebra with you on it. The reason it's so small is because that unit is dealing with it in seconds, rather than minutes or hours.

Answer 34

So you got the right first digits, heh, but you don't have it to the appropriate power. Do you understand what I did math-wise?

Answer 35

yea! lol im taking note here too

Answer 36

I think she went to bed :/

I hope I figured it out in time for her to get it.

The constant changes depending on whether you choose to relate everything by seconds, minutes, or hours, but your answer should adjust accordingly to equal the same time.

Answer 37

yea i think she went to bed lol but good explanation