# Newton's law of cooling/warming

• Sep 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.
• Sep 22nd 2009, 10:36 AM
mikegar813
22.46°c?
• Sep 22nd 2009, 03:03 PM
mr fantastic
Quote:

Originally Posted by mikegar813
22.46°c?

• Sep 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?
• Sep 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?

• Sep 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.
• Sep 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 \$\displaystyle H_0= 4\$, \$\displaystyle H_s= 30\$ and your formula becomes \$\displaystyle 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: \$\displaystyle 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.
• Sep 23rd 2009, 06:30 AM
mikegar813
so 'k' would not change even though the 't' time does?
• Sep 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.