Newton's Law of Cooling
Date: 05/06/2003 at 18:45:54 From: Jenn Subject: Newton's Law of Cooling A famous math professor is found dead by his butler in his 50-room mansion. A student detective arrives on the scene and immediately measures the temperature of the body to be 87 deg F. After 3 hours the student measures the temperature of the body again and finds it has fallen to 84 deg F. Assuming the professor's original temperature was 98.6 deg F, how long before the student arrived on the scene did the professor die? (I-F)e^-kt+F (98.6-?)e^-kt+F
Date: 05/06/2003 at 19:33:59 From: Doctor Luis Subject: Re: Newton's Law of Cooling Hi Jenn, You almost have it with your two formulas (I-F)e^-kt+F (98.6-?)e^-kt+F but you need to organize them better because they were taken at different times. Newton's law of cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and the environment. So, suppose the temperature of the professor after being killed is P(t), where t=0 is the time he was killed. dP/dt = k(R-P(t)) In this formula, k is the rate of cooling (in deg F/hour), R is the room temperature, which I'll assume to be 68 degrees F, although it typically varies from 68F to 77F. So, solving for P in the equation above, we find dP/(P(t)-R) = -k dt ln(P(t)-R) = -kt + ln C P(t) = R + C e^(-kt) So far, we don't know the constants C or k. But we do know three things: P(0 hours) = 98.6 degrees F P(T hours) = 87 degrees F P(T+3 hours) = 84 degrees F where T is the time the professor was found by the student. This is also an unknown. This gives us three equations that we can use to find our three unknowns. They are, 98.6 = R + C 87 = R + C e^(kT) 84 = R + C e^(k(T+3)) = R + C e^(kT) e^(3k) Recall that R = 68. The first equation gives us the value of C. We can use the second equation, plus our new knowledge of C, to find the value of kT. And finally, we can use our knowledge of kT and C in the third equation to find the value of k. Once you know k, you can use the value of kT to find T. The professor died about 8 hours 18 minutes before he was found (T=8.319). What is the value of C and k? I guess I should add that if R = 77 degrees F, then solving for T again, you'd find that T = 6.477 hours (about 6 hours 28 minutes). So there's an uncertainty of (8.319-6.477)hours = .921 hours = 55 mins in the time of the professor's death. It would depend on the room temperature over the time he's been dead. Unless you knew the true time dependence R(t) of the room temperature and solving the differential equation again, you're probably better off making a high and a low estimate of R.
Date: 05/06/2003 at 19:51:00 From: Doctor Anthony Subject: Re: Newton's Law of Cooling If the ambient temperature of the room is @, then Newton's Law of Cooling gives the equation dT/dt = -k.(T - @) where T = temperature at time t. dT ----- = -k.dt and integrating T-@ ln(T-@) = -kt + C T-@ = e^(-kt+C) T-@ = A.e(-kt) at t=0 T = 87 87-@ = A T-@ = (87-@).e^(-kt) at t=3 T = 84 84-@ = (87-@).[e^(-k)]^3 (84-@) ------ = [e^(-k)]^3 87-@ so e^(-k) = [(84-@)/(87-@)]^(1/3) When t = 98.6 we have 98.6-@ = (87-@).e^(-k)^t 98.6-@ = (87-@).[(84-@)/(87-@)]^(t/3) We can solve this if we know the ambient temperature @. If we assume the standard ambient temperature of 60 degrees F then we have 38.6 = 27(24/27)^(t/3) 1.42963 = (8/9)^(t/3) Taking logs we get (t/3)ln(8/9) = ln(1.42963) 3.ln(1.42963) t = ------------- ln(8/9) = -9.1035 = 9 hrs 6 mins 13 secs So the professor died 9 hours 6 minutes before he was discovered by the student. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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