Body Temperature Lab Report

I had gotten a call on Monday at around 4:00 in the morning to come in at the large firm as a dead body had been found in the building. I reached at the large firm at 5:00 AM and found police cars, ambulances, and fire trucks at the scene. A few of my colleagues had come up to me to give me the information from the cleaning crew who had found the dead body. I asked the coroner to measure the body’s temperature and he told me it was currently 88.2oF. I decided to wait an hour to see the decrease in the body’s temperature. Until then I decided to search for more evidence. One of the cleaners told me the building stays at a constant temperature of 67oF. At 6:00 AM, the coroner measured the body’s temperature and it had decreased 1.5 degrees. The body’s temperature was

So far the equation is 19.2 = 21.2e-k. However, k is still not by itself, so I divided 21.2 from each side and ended up with .9292453 = e-k. To get rid of the e, I had to use the inverse, ln(x), on both sides. I had ln(.9292453) = -.0733825, which is k. So following the order of the formula, the equation was B(t) = 67 + 21.2e(-.07t). Now I had to solve for t, so the formula was B(t) = Ts + (T1-Ts)e(-kt). However, this time B(t) = 88.2 and T1 = 98.6 because I was finding the time of death and assumed the person’s body was at normal temperature when he died. The equation is 88.2 = 67 + (98.6 – 67)e(-.07t). I had to isolate the t, so again I subtracted what was in the parenthesis and then subtracted 67 from both sides. After that, the equation was 21.2 = 31.6e(-.07t). Then, I divided 31.6 from each side and ended with .670886076 = e(-.0733825t). Since I had to get t by itself, I had to take the inverse of e on both sides, which is ln(x) and that gave me -.3991559 = -.0733825t. To get t by itself, I divided by -.0733825 on both sides and t = 5.45089088. To round to decimal places that gives me 5.5 which is about 11:30 PM the previous day. The man had died 11:30 PM on