Question

Using Newton's Law of Cooling. You are a medical examiner called to an abandoned house in which a dead body has been discovered. Upon arrival at 1:52 AM, you take the temperature of the body (87 deg) and note that the temperature in the house is a chilly 52 deg.
a) Being well trained in differential equations, you know your first step is to fill in 52 deg into the differential equation, and then distribute the α .
b) Next, you find the general solution to the differential equation. (This will involve α since it is still to be determined).
c) You decide that t=0 will correspond to when you first arrive at the house. You fill in the initial temperature and solve for C.
d) By this time, because your algebra skills are rusty, 2 minutes have elapsed. You are still waiting for the chief of police to arrive, so you decide to waste time by checking your twitter feed. Eight minutes additional minutes later, you tear yourself away and take another temperature of the body. Noting that the body has now cooled to 84 deg, you substitute the new time and temperature into the solution. This allows you to solve for α .
Because you pride yourself on your accuracy, you find α to 4 decimal places.
e) Finally, as you hear the chief of police’s car pulling up, you substitute the temperature of a health living person (98.6 deg) into the equation and solve for t. Because t is negative, you work backwards from the time of your arrival, and say confidently to the chief of police as she arrives: “The murder took place at exactly_______.”

Answer 1

A) The differential equation comes from the fact that the rate of temperature change is proportional to the difference in temperatures.
$\frac{dT}{dt} = \alpha(T -52)$

B) Find general solution by separating variables and integrating$\int\frac{dT}{T-52} = \alpha \int dt \\ ln (T-52) = \alpha t + C \\ T = C e^{\alpha t} +52$:


C) Initial condition is t=0, T = 87
$87 = C +52 \\ C = 35$

D) Total time elapsed is 10 minutes, new temperature is 84
$84 = 35 e^{10\alpha} +52$
solve for alpha
$e^{10\alpha} =\frac{32}{35} \\ \alpha = \frac{ln(32/35)}{10} \\ \alpha = -.00896$

E) Temperature function is:
$T = 35 e^{-.00896 t} +52$
solving for t
$t = \frac{ln (\frac{T-52}{35})}{-.00896}$
Plug in T = 98.6
$t = -31.95$

This is approximately 32 minutes before he arrived or about 1:20 AM