Calculating Time of Death Using Newton's Law of Cooling

rocomath
Messages
1,752
Reaction score
1
So I have a slight problem since I'm getting a negative for time.

A murder victim is discovered at midnight and the temperature of the body is recorded at 31C. One hour later, the temperature of the body is 29C. Assume that the surrounding air temperature remains constant at 21C.

a) Find k

T=A+(T_0-A)e^{-kt}

T(1)=21+(31-21)e^{-k}=29

k=\ln{\frac 5 4}

b) Solve for t

37=21+(31-21)e^{-t\ln{\frac 5 4}}

16=10e^{\ln{\frac 4 5}^t}

t=\frac{\ln{\frac 8 5}}{\ln{\frac 4 5}}

But this gives me a negative t. Would that be okay since I'm trying to use this t value to compute an earlier time?
 
Physics news on Phys.org
Yes.

t=0 when the body is discovered and is at 31 C. So it would have been 37 C at an earlier (negative) time.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top