Newton's law of cooling/warming

Printable View

September 22nd 2009, 10:28 AM
mikegar813

Newton's law of cooling/warming

A thermometer reading 4˚C is brought into a room where the temperature reading is 30˚C. If the thermometer reads 10˚C after 2 minutes, determine the temperature reading 5 minutes after the thermometer is first brought into the room.
September 22nd 2009, 10:36 AM
mikegar813

22.46°c?
September 22nd 2009, 03:03 PM
mr fantastic

Quote:

Originally Posted by mikegar813

22.46°c?

If you show the work leading to your answer it will be easier to check things.
September 23rd 2009, 03:24 AM
mikegar813

30+(4-30)e^(-k2)
k=-.5ln (10/13)
30+(-26)e^((-3)-.5ln (10/13))=12.46
2 minutes = 10°C
3 minutes = 12.46°C
10+12.46= 22.46°C

is this correct?
September 23rd 2009, 04:00 AM
mr fantastic

Quote:

Originally Posted by mikegar813

30+(4-30)e^(-k2)
k=-.5ln (10/13)
30+(-26)e^((-3)-.5ln (10/13))=12.46
2 minutes = 10°C
3 minutes = 12.46°C
10+12.46= 22.46°C

is this correct?

I have no idea what this all means. Where have you used Newton's law of cooling/warming to set up the differential equation? Where is your working for solving that differential equation?

Please show ALL the work leading to your answer if you want your answer reviewed.
September 23rd 2009, 04:21 AM
mikegar813

H= temperature of the object at time t
H sub s= the constant surrounding temperature
t= time
H sub 0= the temperature at t=0
H-H sub s=(H sub 0-H sub s)e^(-kt)

I found this formula in my book.
September 23rd 2009, 05:25 AM
HallsofIvy

So you didn't actually use "Newton' law of cooling", you used a formula derived from that.

Okay, here $H_0= 4$ , $H_s= 30$ and your formula becomes $H= (4- 30)e^{kt}$ . You want to find H when t= 5 but you still don't know what "k" is. You can find that by using the fact that H= 10 when t= 2: $10= (4- 30)e^{2k}$ . Then you can find H when t= 5 by using that same k and t= 5.

Warning: This is NOT linear! H(5) is NOT equal to H(2)+ H(3)! Do NOT find H(3), just use the formula with t= 5, not 3.
September 23rd 2009, 06:30 AM
mikegar813

so 'k' would not change even though the 't' time does?
September 23rd 2009, 03:10 PM
mr fantastic

Quote:

Originally Posted by mikegar813

so 'k' would not change even though the 't' time does?

k is a constant. It does not change.

Were you expected to use Newton's Law of Cooling to solve this question? If yes, then the fact that k is a constant should be clear and you shouldn't be taking shortcuts by applying formulas you don't understand. If not, then the title of your post is misleading and a lot of time might have been wasted answering this question in a way that would ultimately have made no sense to you.