# Newton's Law of Cooling (Calculating Time of Death)

1. May 1, 2012

### JacksonSolo

1. The problem statement, all variables and given/known data

The problem states that you discovered the body at 1pm Thursday in a freezer where the temperature was 10F. Temperature of the corpse at discovery was 40F. I have to find how many hours ago the victim died.

2. Relevant equations

You are given the formula T = Ta + (98.6 - Ta)(0.97)^t where Ta is air temperature.

3. The attempt at a solution

So I plugged the numbers in: 40 = 10 + (98.6-10)(0.97)^t but I have no idea how to solve for t. Any help?

2. May 1, 2012

### sid9221

Move the constants to one side and take logs on both sides you should get:

$$t log(0.97) = log \frac{30}{88.6}$$

Evaluate that for t.

3. May 1, 2012

### JacksonSolo

Ok so I get the log of 30/88.6 and divide that by the log of 0.97 to get t, which equals 35.55 hours. Did I do that right?

4. May 1, 2012

### JacksonSolo

Im just looking for confirmation i solved the problem right.

5. May 1, 2012

### sharks

It appears correct to me. But the answer is actually... 1:45 am on Wednesday?

Last edited: May 1, 2012
6. May 1, 2012

### JacksonSolo

Ok thanks. Yes, true ;)