Newtons law of cooling

At time t=0, T=100 which cools down to T=75 in 2 min,then further I need to find T at t=4,5,8.The issue is I am not given cooling constant K and it is not to be assumed 1/min Equation is dT/dt = k*(T-T_0) where T_0 is 30

[Torsten]

Hint:

You can compute k from the condition that it cools down from 100 to 75 in 2 min. This can be done either numerically or - since the general solution of this simple ODE is known - analytically.

Answer 1

[Sulaymon Eshkabilov]

Is this what you want to compute:

% Given intial data:
T0 = 30;                
T_initial = 100;     
T_target = 75;      
time_step = 2;  

% How to calculate the constant k:
syms k t
EQN = T_target==((T_initial - T0)) *exp(k*t)+T0;
k = solve(EQN, k)

k = 

t=2;
k=subs(k, t);
k=double(k)

k = -0.2209

% Diff EQN:
dT = @(T) k * (T - T0);

% Solve the Diff EQN @ t = 4, 5, 8:
t_vals = [4, 5, 8]; 
for i = 1:length(t_vals)
    t = t_vals(i);
    % Solution T at given time t:
    [time, T] = ode45(@(t, T) dT(T), [0 t], T_initial);
    fprintf('At t = %d minutes, T = %.2f\n', t, T(end));
end

At t = 4 minutes, T = 58.93
At t = 5 minutes, T = 53.19
At t = 8 minutes, T = 41.96

If you want more detailed calcs with smaller steps

% Solve the Diff EQN @ t = 4, 5, 8:
t_vals = 2:.5:10; 
for i = 1:length(t_vals)
    t = t_vals(i);
    % Solution T at given time t:
    [time, T] = ode45(@(t, T) dT(T), [0 t], T_initial);
    fprintf('At t = %d minutes, T = %.2f\n', t, T(end));
end

At t = 2 minutes, T = 75.00
At t = 2.500000e+00 minutes, T = 70.29
At t = 3 minutes, T = 66.08
At t = 3.500000e+00 minutes, T = 62.31
At t = 4 minutes, T = 58.93
At t = 4.500000e+00 minutes, T = 55.90
At t = 5 minutes, T = 53.19
At t = 5.500000e+00 minutes, T = 50.77
At t = 6 minutes, T = 48.60
At t = 6.500000e+00 minutes, T = 46.65
At t = 7 minutes, T = 44.91
At t = 7.500000e+00 minutes, T = 43.35
At t = 8 minutes, T = 41.96
At t = 8.500000e+00 minutes, T = 40.70
At t = 9 minutes, T = 39.59
At t = 9.500000e+00 minutes, T = 38.58
At t = 10 minutes, T = 37.69

plot(time, T)

[John D'Errico]

Please don't do obvious homework assignments for students who have not shown the slightest effort. That does not help the student, and it actively hurts the the site, since now this student will think they can keep on doing the same, and other students will think the same.

[Habiba]

Firstly,I have already tried and I haven't asked you to solve entire problem just asking for hint to find k. Secondly, this is not included in my university course, my teacher has given us assignment without teaching us and most of my other class fellows are not even making the effort I am making.

[Torsten]

@Habiba

Why did you accept the wrong solution from @Sulaymon Eshkabilov and not the correct one given by @Sam Chak ? Don't you see that T(2 min) = 78.98 instead of 75 for

k = (1 / (T_initial - T0)) * ((T_target - T_initial) / time_step);

?

[Dyuman Joshi]

@Habiba, John D'Errico's comment is not directed at you, but the person who has answered it.

And if you have made an effort to solve the problem, it is better to share that with the question.

[Habiba]

@dyuman Don't you have any other work except poking your nose into others matter. I have accepted the answer that helped me more and also voted to sam chak who had made an effort. I don't know that the code is giving wrong answer because I have made my own code not just copied it.

[Dyuman Joshi]

Lol.

Answer 2

[Sam Chak]

Hi @Habiba

I understand that you are grappling with a certain problem. That's why you've outlined the challenge of not knowing the rate of heat loss, denoted as k. More importantly, you haven't explicitly requested us to solve the problem for you, and this is greatly appreciated. It suggests that you have a keen interest in acquiring knowledge and derive satisfaction from the process of learning how to solve problems.

People perceive and grasp mathematical concepts differently. A math teacher views the world of differential equations through a distinct lens, as they stand on the shoulders of Newton, while a student may be positioned at the feet of Newton.

For instance, if you possess the key to solving first-order linear ordinary differential equations, wherein the solution to

is expressed as (according to the Schaum's Outline of Mathematical Handbook of Formulas and Tables)

,

then you can determine the value of k such that .

By now, you probably have noticed that at , . Since and are known, can you solve the algebraic equation and determine k?

[Habiba]

Yes, Thankyou, now I get it.

[Sam Chak]

@Habiba, good to hear that! If you find the proposed approach helpful, please consider clicking 'Accept' ✔ on the answer and voting 👍 for it. Thanks a bunch! By the way, you can also vote for other helpful answer as a token of appreciation.