# Thread: Another Newton's Law of Cooling/Warming

1. ## Another Newton's Law of Cooling/Warming

A small metal bar, whose initial temp was 20C, is dropped into a large container of boiling water. How long will it take the bar to reach 90C if it is known that its temp increases 2 degrees in 1 second? How long will it take it to reach 98C?

With this problem, I'm not sure where to even begin, other than with the equation

dT/dt = k(T-Tm)

What is the initial temperature, 20C at T(0)? How does 2 degrees in 1 second play into it? How do I find k? The book gives no examples, so I'm not sure how to even start.

--Hank

2. ## Re: Another Newton's Law of Cooling/Warming

Originally Posted by hank
A small metal bar, whose initial temp was 20C, is dropped into a large container of boiling water. How long will it take the bar to reach 90C if it is known that its temp increases 2 degrees in 1 second? How long will it take it to reach 98C?

With this problem, I'm not sure where to even begin, other than with the equation

dT/dt = k(T-Tm)

What is the initial temperature, 20C at T(0)? How does 2 degrees in 1 second play into it? How do I find k? The book gives no examples, so I'm not sure how to even start.

--Hank
Please check the problem carefully for accuracy. By any cance did the problem say

"...that its temp initially increases 2 degrees in 1 second..."

3. ## Re: Another Newton's Law of Cooling/Warming

Originally Posted by Subhotosh Khan
Please check the problem carefully for accuracy. By any cance did the problem say

"...that its temp initially increases 2 degrees in 1 second..."
Thanks for getting back to me.
I double checked the problem, and what I typed is correct.
Perhaps the book made a typo?

The answers given in the back of the book are 82.1s and 145.7s.

4. ## Re: Another Newton's Law of Cooling/Warming

i have the same problem.
I think I solved it, but the answer isn't exact.
I keep getting my rate as 0.025
I know that's the problem but I can't seem to find another rate.
If i modify the rate to 0.02531 it works out well.
thx

5. ## Re: Another Newton's Law of Cooling/Warming

However, if the problem is correctly stated (as posted), then

dT/dt = 2 = constant

so

T = T[sub:34llk5uc]o[/sub:34llk5uc] + 2 *t

T = 20 + 2*t

However, that is highly unlikely

6. ## Re: Another Newton's Law of Cooling/Warming

the problem is stated correctly.

here is what I did, (again there is something wrong with my rate, what should I do differently?)
dT/dt = k(T-A)
2 = k(20 -100)
k = -0.025

*it is heating up instead of cooling.*

T(t)= T + (A-T)e^(kt)
T(t)= 100 + (20 -100)e^(kt)
T(0)= 100 -80
T(0)= 20
(this makes sense thus far, the temperature of the metal bar when t=0 is 20)

*for the first temperature*
90= 100 + (20 -100)e^(-0.025t)
-10= -80e^(-0.025t)
Ln(1/8)= -0.025t
t = 83.17 s (answer is supposed to be 82.1s)

*for the second temperature*
98= 100 + (20 -100)e^(-0.025t)
-2=-80e^(-0.025t)
0.025=e^(-0.025t)
Ln(0.025)= -0.025t
t = 147.55 s (answer is supposed to be 145 s)

The error has to be with my rate (k). If I change 0.025 to 0.02531 it works out to the correct answer.
How can I obtain this different rate?
any help is appreciated.

7. ## Re: Another Newton's Law of Cooling/Warming

Originally Posted by sifusith
the problem is stated correctly.

here is what I did, (again there is something wrong with my rate, what should I do differently?)
dT/dt = k(T-A)
2 = k(20 -100)<<< You are assuming that the initial rate is 2 °C/min - the problem does not say so.

8. ## Re: Another Newton's Law of Cooling/Warming

Originally Posted by hank
A small metal bar, whose initial temp was 20C, is dropped into a large container of boiling water. How long will it take the bar to reach 90C if its temp increases 2 degrees in 1 second?
Am I misinterpreting the given condition?
I don't know how to find the rate.

9. ## Re: Another Newton's Law of Cooling/Warming

Your interpretation would be correct - if the problem said:

How long will it take the bar to reach 90C if initially its temp increases 2 degrees in 1 second

10. ## Re: Another Newton's Law of Cooling/Warming

ok, I see.
but I still don't know how to find the rate.
what do I use the 2C in 1 sec for?
how should I go about finding the rate?

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•