forensics uses newton's law of cooling to determine when a victim died.
T = temperature
t = time
Rt = room temperature
To = initial temp.
k = rate of cooling.
the equation is T = (To - Rt)e^kt + Rt
you would use logs to solve the equation.
For example, a victim was found in a room that had been kept at a constant temperature of 68 degrees farenheit for three days. Upon arrival, at 10:23 pm, the temperature of the body was taken and found to be 80 degrees farenheit. In an hour, the temperature is taken again and found to be 78.5 degrees farenheit. When did the victim die?
to find this answer, we can set up a table (read the coordinate up and down)
t (hours)-----------------> 0 ..... 1
T (temp in degrees F) 80 ... 78.5
so, we can set the equation like this: T = (80-68)e^(kt) + 68
then, we can use the other point we have to get: 78.5 = 12e^(1k) + 68
subtract 68 from both sides to get 10.5, divide by twelve to get: 7/8 = e^k
then, we use logs. ln(7/8) = lne^k
therefore: ln(7/8) = k which is equal to about -0.1335.
assuming the victim had normal body temperature, 98.6 degrees farenheit, we can substitute this in the equation: 98.6 = 12e^(0.1335t) + 68
subtract 68 and divide by 12 to get 2.55 = e^(-0.1335t)
ln(2.55) = lne^(0.1335t); therefore, ln(2.55) = -0.1335t
divide ln(2.55) by -0.1335 to get t = -7.01
therefore, time of death is equal to 7.01 hours before the forensic team arrived at 10:23.
by the use of natural logs and exponential equations, they can find that the victim's death was aprroximately 3:22 pm that same day.
T = temperature
t = time
Rt = room temperature
To = initial temp.
k = rate of cooling.
the equation is T = (To - Rt)e^kt + Rt
you would use logs to solve the equation.
For example, a victim was found in a room that had been kept at a constant temperature of 68 degrees farenheit for three days. Upon arrival, at 10:23 pm, the temperature of the body was taken and found to be 80 degrees farenheit. In an hour, the temperature is taken again and found to be 78.5 degrees farenheit. When did the victim die?
to find this answer, we can set up a table (read the coordinate up and down)
t (hours)-----------------> 0 ..... 1
T (temp in degrees F) 80 ... 78.5
so, we can set the equation like this: T = (80-68)e^(kt) + 68
then, we can use the other point we have to get: 78.5 = 12e^(1k) + 68
subtract 68 from both sides to get 10.5, divide by twelve to get: 7/8 = e^k
then, we use logs. ln(7/8) = lne^k
therefore: ln(7/8) = k which is equal to about -0.1335.
assuming the victim had normal body temperature, 98.6 degrees farenheit, we can substitute this in the equation: 98.6 = 12e^(0.1335t) + 68
subtract 68 and divide by 12 to get 2.55 = e^(-0.1335t)
ln(2.55) = lne^(0.1335t); therefore, ln(2.55) = -0.1335t
divide ln(2.55) by -0.1335 to get t = -7.01
therefore, time of death is equal to 7.01 hours before the forensic team arrived at 10:23.
by the use of natural logs and exponential equations, they can find that the victim's death was aprroximately 3:22 pm that same day.
-
at the bank.. when you're calculating the interest in your account.