# Newton's Law of Cooling Watch

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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?

could someone please give me the formula(s) for Newton's Law of Cooling and explain how to use it?

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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

And I don't understand how to really use it.

I am getting several different formulas from the internet

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#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

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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#5

T = A e^-kt

Where T = instananeous Temperature

A = Intial temperature

k = decay constant

t= time

Where T = instananeous Temperature

A = Intial temperature

k = decay constant

t= time

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How do I find C (integration constant?) Could someone please explain because I have a test tomorrow.

thanks

thanks

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#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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#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.

(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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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?

Reply ASAP please

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?

Reply ASAP please

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#10

(Original post by

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?

Reply ASAP please

**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?

Reply ASAP please

and use k in the formula to obtain... 2 = 28 e^-kt... to find t

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