differential equations: heat transfer?

Question

the body of a murder victim was discovered at 11:00 am. the medical examiner arrived at 11:30 am and found the temperature of the body was 94.6 degrees F. the temperature of the room was 70 degrees F. one hour later, in the same room, he took the temp. again and found that it was 93.4 degrees F. estimate the time of death.

please tell me how you figured out your answer!

standard body temperature is 98.6 degrees F

Answer 1

tell me what is normal t°F for a healthy human, I know t°C=36.6°; I don’t know Fahr scale! 98.6°F – thank you.
♠ one of Newton’s laws says: dT° =-k*(T(t)-Tr)*dt, where dT is change of body temperature during a short time dt, Tr is constant temperature of the room, T(t) is temperature of the body at a moment t, while k is constant to be determined, sign minus showing that the body is cooling;
♣ thus dT/(T-Tr) =-k*dt; integrating we receive: ln((T-Tr)/(T0-Tr)) = -k*t, where T0 =98.6°F, Tr=70°F; or T-Tr = (T0-Tr)*e^(-k*t), k is rate to bi determined;
♦ at t1=11:30 T(t1)=94.6°, so 94.6-70 = (98.6-70)*e^(-k*(t1-t0));
♥ at t2=12:30 T(t1)=93.4°, so 93.4-70 = 28.6*e^(-k*(t2-t0)); t2-t1=1; t0 is time of murder;
dividing ♦/♥: 24.6/23.4 =e^(k*1), hence k=ln(24.6/23.4) = 0.05;
return to ♦; 24.6 = 28.6*e^(-0.05*(t1-t0)); or ln(24.6/28.6) =-0.05(t1-t0),
hence t1-t0=3,01 hours back; the murder was committed at t0 =11:30 –3 =8:30;

Answer 2

I have no idea if this is right...

in one hour, the body dropped 1.2 degrees. (94.6 - 93.4)

At 11:30 the body was 94.6, which is 4 degrees less than normal body temp (98.6).

So... 4/1.2 = 3 1/3 hours (3 hours, 20 minutes) since death.

11:30 - 3 hr 20 min = 8:10.

Answer 3

The loss of heat is through convection, which depends on the difference between the body temp and the air temp. The differential equation governing this has an exponential decay form of solution.

So, what you're being asked to do is, use this equation,
where
Ts is the starting temp of the body at t=0
Ta is the fixed air temp
Tb(t) is the bodies Temp vs time

Tb(t) = (Ts-Ta)*exp(-kt) + Ta

use the two data points to get k, the decay constant, then Ts

Note that what this means is that the body's temperature asymtotically approaches the air temperature.

Btw, the differential equation for convective heat transfer is

dT/dt = -k(T-Ta)

Btw, I taught heat transfer at a university and worked in spacecraft thermal design for a couple of years.

Answer 4

so in 1hr....it dropped...1.2degrees F. rite? so divide that in half cuz it's only 30 mins after he died that the doc took his temp. so .6

Answer 5

x'+x/t1=n*H(t) The integrating ingredient is then exp(int(one million/t1))=exp(t/t1) multiply via: exp(t/t1)*x'+x/t1*exp(t/t1)=n*H(t)*exp..... simplifies to: d(exp(t/t1)*x)/dt=n*H(t)*exp(t/t1) because of the fact that H(t) is slightly clever function, we ought to evaluate this in 2 areas, one the place that's h, and one the place's it 0, so we get H=h: exp(t/t1)*x+C1=int(n*h*exp(t/t1),dt) exp(t/t1)*x+C1=n*h*int(exp(t/t1),dt)=n..... x=n*h*t1+C1*exp(-t/t1) H=0 exp(t/t1)*x+C2=int(n*0*exp(t/t1),dt) exp(t/t1)*x=C2 x=C2*exp(-t/t1) regrettably, the constants must be evaluated every time, a area harm variations, so: t=0: x=n*h*t1+C1*exp(-t/t1) @t=0, x=.25*sixteen*5+C1*exp(0) C1=-3 then @t=L x1=20-3*exp(L/5) so: x2=20-3*exp(L/5)=C2*exp(-L/5) C2=20*exp(L/5)-3*exp(2*L/5) then we could repeat at t=one million, and t=one million+L, and proceed on...