## anonymous 4 years ago 1. Newton’s law of cooling states that the rate of cooling of an object is proportional to the difference between its temperature and the ambient temperature. (a) Assuming this, formulate an initial value problem that models the cooling of a cup of coffee. (b) Suppose that the ambient temperature is 20◦ C, the coffee is initially at a temperature of 90◦ C, and is initially cooling at a rate of 8◦/minute. How long will it take for the coffee to cool to 75◦.

1. anonymous

i believe the answer is1.875 minutes

2. anonymous

can someone instruct me how to get the answer ...

3. JamesJ

Let $$T_A$$ be the ambient temperature and T the temperature of the object cooling. What the problem tells you is $\frac{dT}{dt} \ \ \ \alpha \ \ - (T - T_A)$ hence you can write down the ODE with a constant of proportionality.

4. JamesJ

Now. Use the facts given to you in part (b) to calculate that constant. Then, solve the ODE and apply the initial values given to obtain the solution to this problem. Finally, find the time t such that T = 75 C.

5. JamesJ

So, does what I wrote down above make sense and do you see what to do next?

6. anonymous

not really .. i dont think that we're supposed to use the second part of the question to figure out part a.. and i'm not sure where the alpha came from

7. JamesJ

alpha means "is proportional to"

8. JamesJ

The change in temperature of the object, like a cup of hot water, is proportional to the difference of its temperature to the ambient temperature. The negative sign is there because the temperature is going down towards the ambient temperature if it currently above it; or changing upwards, if T < T_a.

9. anonymous

hmmm.. i wish i understood, thanks anyways though

10. JamesJ

Let's try this one more time. You're told that: "Newton’s law of cooling states that the rate of cooling of an object is proportional to the difference between its temperature and the ambient temperature." Let's write T(t) for the temperature of the cooling object. This law of cooling is telling us something about $\frac{dT}{dt}$ so far so good?

11. anonymous

ok.

12. anonymous

ok i think i may have misread the question ... it's asking for an inital problem and not initial formula .. i'm not really sure what it's asking.

13. JamesJ

Now, what is the change proportional to? "proportional to the difference between its temperature and the ambient temperature" Let $$T_A$$ be the ambient temperature, then the difference between the temperature T and the ambient temperature is $T - T_A.$ Hence $\frac{dT}{dt} \ \text{ is proportional to } -(T-T_A)$ (We write $\frac{dT}{dt} \ \ \alpha \ -(T-T_A)$ this is standard notation; I'm sure you've seen it before, but it isn't very important for this problem.)

14. JamesJ

What is important is the negative sign. It must be negative because if $$T > T_A$$ then T should be going down. I.e., the object is cooling and its temperature T should be getting closer to the ambient temperature. So far, so good? Be honest.

15. anonymous

Ok if I'm being honest, I just started a differential calculus class and I don't even know what we're learning .. I'm not even sure what the objective of the course is. I sound stupid, but promise just takes a little time ahah .. Ok, so for the first part all we're trying to say is that the 2 are proportional to eachother? we dont need an equal sign or anything then in the answer right?

16. JamesJ

We do. We now introduce a constant of proportionality. What that equation means in concrete terms is $\frac{dT}{dt} = -k(T-T_A)$ for some positive constant k. This is the ODE we want.

17. anonymous

ODE means original differential equation?

18. JamesJ

ordinary differential equation.

19. anonymous

lmao alright. you must think i'm a retard.

20. JamesJ

I'm going to refer you now to this lecture where this equation is solved. http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/ I *strongly* recommend you watch it, take notes, rewind, watch it again. We all take a while to understand new concepts. This example of Newton's Law of Cooling is a classic.

21. anonymous

ok thank you.

22. JamesJ

after you've watched it, if you still have doubts, come find me.

23. anonymous

ok .. will do, thanks

24. JamesJ

btw, the solution method he shows you in the lecture is more powerful than what you need. But no harm done, because it is a good general method. I can still show you a simpler method for you equation in particular: separation of variables. You may know it already.

25. JamesJ

but more than "it's a good general method", this is going to be the next thing you learn in your class anyway.

26. anonymous

kind of... i'm working on b now, and for solving ... are you able to walk me through the steps to getting the solution?

27. anonymous

if your still there ... i got T=Ta +_ e^(-kt+c)

28. JamesJ

Right, $T(t) = T_a + Ae^{-kt}$ For some constant A ( = e^c ) Now, in part b, we can solve for A. We have that T(0) = 90. Therefore $90 = T(0) = T_a + A$ hence, $A = 90 - T_a$ We also know that $$T_a = 20$$, thus $$A = 90 - 20 = 70$$ and the solution is $T(t) = 20 + 70e^{-kt}$ ok?

29. anonymous

yes thank you so much!

30. JamesJ

Now, one more thing. You need to find k. How are yo going to do that? You're given one more piece of information in (b) "[the coffee] is initially cooling at a rate of 8◦/minute". Translate that into the mathematics.

31. anonymous

so in math terms thats .. -8t?

32. JamesJ

"Cooling at a rate of 8 degrees/minute" That must mean dT/dt at t = 0 is 8 degrees/minute Now dT/dt = k(T_a - T) You're told that for t = 0 dT/dt = -8 degrees/minute We also have that k(T_a - T) = k(20-90) = -70k Thus -70k = -8 You can now solve for k.

33. anonymous

ohhh .. ok, and then once i've solved for k i can solve for t?

34. JamesJ

Yes, then finally you can find the t for which T(t) = 75

35. anonymous

aren't you just wonderful, thank you so so so much!