MatLab codes for the used by forensics to determine the exact time of death.

Hello:

Please help me write MatLab codes for the solution of the model used by forensics to determine the exact time of death derived from Newton's law of cooling given that T(t)= 70+10e^kt and T(0)=80. I need it for my project and am not so familiar with MATLAB

Those are the only informations i was given

Here I was trying but i do not really have a clue on how to continue

clc; clear ; close all;

%T(t)=70+10e^kt T(0)=80

F = @(t,T(t)) 70+10*exp(k*t);

T_0 = 80;

plot(t,T_0)

Thank you for your help.

Answer 1

[Davide Masiello]

if you already have an expression for T, why would you solve the ODE?

anyways, the correct syntax is

clear,clc
k = 0.5

k = 0.5000

F = @(t,T) k*(70-T);
tspan = [0 5];
T0 = 80;

[t,T] = ode45(F,tspan,T0);
plot(t,T)
legend('k = 0.5')

You could even solve it analytically

syms t T(t) k

eqn = diff(T,t) == k*(70-T);
sol = dsolve(eqn,T(0)==80)

sol = 

%Solution for several values of k
for n = 0.1:0.2:1
solk = subs(sol,k=n);
fplot(solk,[0 5]),hold on
end
legend([repmat('k = ',5,1),num2str((0.1:0.1:0.5)')],'Location','best')

[Evalistus Simon]

I thank you so much but I think something is wrong as the model is based on the Newton's Law of cooling. I mean as time increases is the temperature not supposed to be decreasing? Thnk you

[Davide Masiello]

I think the problem is in the sign of the equation. If the body is cooling it means that the heat flux is negative, hence

with in your case.

I have edited my answer accordingly.

[Evalistus Simon]

Thank you very much.

[Davide Masiello]

My pleasure. If the answer was helpful, don't forget to Accept it!

Answer 2

[Sam Chak]

Hi @Evalistus Simon

The equation derived from Newton's Law of Cooling is given by

The ambient temperature is 70°F. Suppose that the coroner measured the body's temperature when they first arrive at the scene at 8:15 AM. Then, an hour later a second temperature is taken.

If the first temperature is 72.5°F, and the second temperature is 72.0°F, then the coroner can solve for the cooling rate k.

k = - log(2.5/2)

k = -0.2231

Substituting this value back to the equation when the 1st temperature is taken, and solve for

fun = @(t) 70 + 10*exp(-0.2231*t) - 72.5

fun = function_handle with value:
    @(t)70+10*exp(-0.2231*t)-72.5

tsol = fsolve(fun, 10)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

tsol = 6.2138

The coroner can postulate that the person died about 6 hours 13 minutes before 8:15 AM, which would be around 2:00 AM.

t = 2:0.01:9;
T = 70 + 10*exp(-0.2231*(t - 2));
plot(t, T), grid on, xlabel('Time (Clock hour)'), ylabel('Body temperature')