A detective is called to the scene of a crime where a dead body has just been found. She arrives on the scene at 10:23 when she takes the temperature of the body which is 82 degrees F. The detective checks the thermostat and finds the room has been kept at 68 degrees F for the past three days.
at 11:23 the temperature of the body is taken again and is at 80.5 degrees F.
The next day the detective is asked 'what time did the victim die' assuming the victims body temp was normal (98.6 degrees F) prior to death, what is the answer to the question?

I don't know what equation to use in order to 'relate' the rates... HELP?(: pleaseeee.

You need to set up and solve the differential equation that governs heat flow from a body into ambient air. You can probably assume that the rate of change of body temp is proportional to the difference between the body temp and ambient:


\frac {d \theta(t)} {dt} =k(\theta - T_f)


where \theta (t) = body temp as a function of time and T_f = room temp. The solution for this equation is:


\theta (t) = (T_i - T_f)e^{-kt} + T_f


where T_i = initial body temp at time t=0, which is 98.6, and T_f is the final body temp after it equalizes with the room temp of 68 degrees. So you have:


\theta (t) = 30.6 e^{-kt} + 68


Let t_1 = the time from death until the first temperature reading, measured in minutes. You can put in the values for \theta at t= t_1 and t = t_1+60, and from that you can determine a values for k and t_1.

Post back with your answer and we'll check it for you.

The victim died at 10:08?. I don't think that's right but that's what I got.

Wait I reworked the problem and got 9:31 are either correct?

No, unfortunately neither are correct. The time of death was quite a bit earlier than that as your intuition probably told you. If the body only cooled by 1.5 degrees during the hour after she arrived at the scene, it seems pretty unlikely that it had cooled a whopping 16.6 degrees in the previous 52 minutes.

Ebaines gave you one way to solve it, but you may find it easier to choose 10:23 (when the detective first arrived) as t=0. Then you can solve for k directly in one step, rather than having to simultaneously solve two equations for k and t_1. Your equation would be

\theta(t)=14e^{-kt}+68

Since that gives you an initial temperature of 82 degrees at t=0 and a final temperature of 68 degrees at t=infinity (as Ebaines explained before). Then you can easily solve for the decay constant, k (no pun intended :)), by plugging in the appropriate numbers for the second data point 60 minutes later:

80.5 = 14e^{-k \cdot 60} + 68

Once you have k, then you can solve for the time of death by plugging in k and solving for t in this equation:

98.6 = 14e^{-kt} + 68

You should get a relatively large (many hundreds of minutes) negative value for t. It's negative because it's the number of minutes before t=0 that the death took place.

Hopefully that helps.