Newton's law of cooling gives the temperature T(t) of an object at time t as where and Consider the following scenario: Use Matlab to do the following: Determine the value of k. Determine the time of death, assuming that victim's body temperature was normal ( 98.6° F) prior to death, and that the room temperature was constant at 69° F. Calculate the temperature of the body for every minute between 9:18 PM and 10:18 PM Plot the temperature of the body for every minute between 9:18 PM and 10:18 PM. Include a plot title and axis labels.

The code as well as the outputs are presented in the parts respectively.

Explanation:

As the question is not complete, the complete question is as attached in the figure.

The matlab code is as given below

Part a:

The Newton's Law is given as

Rearranging this equation gives

or

For this case the values are given as below:

T(10:18pm)= 78 F

T(9:18pm)= 79.5 F

T_s=69 F

t=1 hour

So the value of k is calculated as per following

Code:

%% For estimating the value of k

T_t=78;T_o=79.5;T_s=69;t=1; %initializing the variables

k=-(log((T_t-T_s)/(T_o-T_s)))/t;% Calculating the value of t in hr^-1

disp(['The value of k is ',num2str(k),' 1/hr'])

Output:

The value of k is 0.15415 1/hr

Part b:

For calculation of time of death following data is used

T(9:18pm)= 79.5 F

T(0)=98.6 F

T_s=69 F

k=0.15415 1/hr

Code:

%% For estimating the time of death

T_o=98.6;T_t=79.5;T_s=69; %initializing the variables

t=-(log((T_t-T_s)/(T_o-T_s)))/k;% Calculating the value of t in hr^-1

t_n=(9+12)+(18/60);%converting the time 9:18 pm into 24-hours format

t_i=t_n-t%finding the time of death in 24-hours format

t_ih=floor(t_i)%Converting time to HH:MM format (hours)

t_im=ceil((t_i-t_ih)*60)%Converting time to HH:MM format (mins)

if (t_ih)>=12%Converting time to AM/PM format

   t_ih=t_ih-12;

   s=' PM'

else

   t_ih=t_ih;

   s=' AM'

end

disp(['The time of death is ',num2str(t_ih),':',num2str(t_im),s])

Output:

The time of death is 2:35 PM

Part c:

For calculation of temperature between 9:18 PM and 10:18 PM following data is used

T_0= 79.5 F

T_s=69 F

k=0.15415 1/hr

Code:

%% Calculating value of temperature for each minute between 9:18 PM and 10:18 PM

T_o=79.5;T_s=69; %initializing the variables

t=1/60; %Converting a minute to hour

T=zeros(60,1);%Initializing the array to store the temperature

e=exp(1);

for i=1:1:60%counter for an hour

   T(i,1)=T_s+(T_o-T_s)*e^(-k*t);%Estimating the temperature after 1 min

   T_o=T(i,1);%Changing the T_o

end

disp('The temperature variation with time in mins is as below')

T

Output:

The temperature variation with time in mins is as below

T =

  79.4731

  79.4462

  79.4194

  79.3926

  79.3660

  79.3394

  79.3129

  79.2864

  79.2600

  79.2337

  79.2074

  79.1812

  79.1551

  79.1290

  79.1031

  79.0771

  79.0513

  79.0255

  78.9998

  78.9741

  78.9485

  78.9230

  78.8975

  78.8721

  78.8468

  78.8215

  78.7963

  78.7712

  78.7461

  78.7211

  78.6962

  78.6713

  78.6465

  78.6217

  78.5970

  78.5724

  78.5478

  78.5233

  78.4989

  78.4745

  78.4502

  78.4260

  78.4018

  78.3777

  78.3536

  78.3296

  78.3057

  78.2818

  78.2580

  78.2342

  78.2105

  78.1869

  78.1633

  78.1398

  78.1164

  78.0930

  78.0696

  78.0464

  78.0232

  78.0000

Part d:

The plot is as attached.

Code:

%% Plotting the temperature profile

t=1:1:60;

plot(t,T,'-o')

xlabel('Time between 9:18 PM and 10:18 PM in mins')

ylabel('Temperature of body in F')

title('The variation of body temperature between 9:18 PM & 10:18 PM')

grid on

The output is also attached in the file

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg