Question

How to do a Newton's Law of Cooling question?

Answer 1

The temperature of a roast varies according to Newton's Law of Cooling:\(\frac{dT}{dt} = -k(T-A)\), A is the room temperature, and k is a positive constant.


If a room temperature roast cools from 68°F to 25°F in 5 hours at freezer temperature of 20°F, how long (to the nearest hour) will it take the roast to cool to 21°F?

1

9

12

24

Answer 2

I used separation of variables and got

\(|T-A| = Ce^{-kt}\)

But I'm not sure what to do now

Answer 3

@imqwerty Hey man, if you could help out a bit here I would appreciate it!

Answer 4

\(\color{#0cbb34}{\text{Originally Posted by}}\) @justjm
I used separation of variables and got

\(|T-A| = Ce^{-kt}\)

But I'm not sure what to do now
\(\color{#0cbb34}{\text{End of Quote}}\)
you can remove that mod because T will always be greater than A
cuz the refrigerator can only cool down something to its own temperature, it won't take it below A

T - A = C\(e^{-kt}\)

Answer 5

A is 20\(^o\)F

Answer 6

T - 20 = C\(e^{-kt}\)

Answer 7

initially, the temperature of the roast is 68\(^o\)F
so if you plug t=0 in the above equation you should get T = 68

Answer 8

note that putting t=0 also eliminates 'k' so you're only left with C which you can find out

Answer 9

once you get 'C' you can use the second condition ( T = 25 when t = 5 ) to find k

Answer 10

you might be wondering whether to take 't' in seconds, minutes or hours but it wouldn't matter because based on what unit you take you'll get different values of k

Answer 11

to find the time taken to cool down to 21F, substitute C, k and T = 21, get t
the 't' that you get here will be the time from the beginning (when the roast was at 68\(^o\)F)

Answer 12

Thank you for taking the time to help! I understand now, and somehow my class did not cover it well.

Answer 13

np :)