Newton's Law of CoolingWatch

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Thread starter 14 years ago
#1
I am not sure whether this should be in the physics forums or the math forum but it does deal with differential equations so;

could someone please give me the formula(s) for Newton's Law of Cooling and explain how to use it?
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14 years ago
#2
I think dT/dt = -kT, where T is temperature, k is constant, and t is the time.
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Thread starter 14 years ago
#3
I need it in the form, T=.....
And I don't understand how to really use it.
I am getting several different formulas from the internet
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14 years ago
#4
If I was right, dT/dt = -kT , so u got dT/T = -kdt.
Integral both side u got
lnT = -kt + C
T = e^(-kt+C) = e^C.e^(-kt) = Ae^(-kt)
where A, C are constants
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14 years ago
#5
T = A e^-kt

Where T = instananeous Temperature
A = Intial temperature
k = decay constant
t= time
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Thread starter 14 years ago
#6
How do I find C (integration constant?) Could someone please explain because I have a test tomorrow.
thanks
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14 years ago
#7
You can't find C without an initial condition or similar - i.e. being told that the temperature is such at a certain time, usually t=0. From this you can get C.
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14 years ago
#8
Newton's law of cooling states that the amount of heat that a body loses is proportional to its temperature:

(dT/dt)~-T (minus because it is decreasing)

=>(dT/dt)=-kT

=>INT[(1/T)dT]=INT[(-k)dt]

=>logT=-kt+C

Raising each side to e:

T=e(-kt+C)

The initial condition is given when t=0

=>t=0=>T=(e^C)=initial temperature=Ti

=>T=Ti(e^(-kt))

Now let us set another initial condition. Let us say that the time it takes for a body to lose a half of its heat is th. Now if we let t=th this means that the ratio between the original temperature and the temperature at this time will be exactly a half:

=>(1/2)=(e^(-kth))

Taking logs of both sides:

log(1/2)=-kth

log(1)-log(2)=-kth

-log(2)=-kth

=>k=(log(2)/th)

Where logx is to base e i. e. lnx

=>T=Ti(e^(-(log(2)/th)(t))

Newton.
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Thread starter 14 years ago
#9
I have been given two equations,
T(t) - R = Ce^(kt)

T- Ts = (To - Ts)e^(kt)

how do I use them?

T is temperature,
R is room temperature,
C and k are constants I presume.

Also, here is a question:
A Beer at 48C goes into a freezer which is -4C. 30 min later the beer is 31C. When will it be 2C?

How do I do this?
0
14 years ago
#10
(Original post by himurakenshin)
I have been given two equations,
T(t) - R = Ce^(kt)

T- Ts = (To - Ts)e^(kt)

how do I use them?

T is temperature,
R is room temperature,
C and k are constants I presume.

Also, here is a question:
A Beer at 48C goes into a freezer which is -4C. 30 min later the beer is 31C. When will it be 2C?

How do I do this?
well... do this... 31 = 28 e^-k(30) ... work out the value of k
and use k in the formula to obtain... 2 = 28 e^-kt... to find t
0
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