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 Post subject: Newton’s law of cooling, hard questionPosted: Wed, 9 May 2012 15:15:35 UTC
 S.O.S. Newbie

Joined: Wed, 9 May 2012 15:07:45 UTC
Posts: 1
By Newton’s law of cooling the temperature T of a cooling object as a function of time is

TO - initial temperature,
T1 is ambient temperature (meaning the temperature outside the cooling object)
k is constant
W

-32C was the te,perature when electricity was cut off and heating of a house stopped working. After 2 hours from the power cut the inside temperature was measured to be 19,2 C, and after that the following measurements were done:

2,5 hours: 18,3 oC
3,0 hours: 17,5 oC
3,5 hours: 16,6 oC
4,0 hours: 15,8 oC
4,5 hours: 14,9 oC
5,0 hours: 14,2 oC

Supposing Newton’s cooling law to be valid in this case,

Search:

k
T0 initial temperature (which is the temperature at moment t=0 when the power was cut off). Try also to predict when the inside temperature will be dropped below zero.

______________________________________________________________________________________________________

my attempt:

19.2-(-32)=(T0-32)e-k(2)
14.2-(-32)=(T0-32)e-k(5)

51.2 (T0-32)e-k(2)
46.2 (T0-32)e-k(5)

1.11 e2,5k
k= 0.041103894
k= ln(1.11)/2.5

Substitude k into the first equation

51.2=(T0-32)e-2k
51.2=(T0-32)e-2((ln(1,11)/2,5))

and now ?????????????

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 Post subject: Re: Newton’s law of cooling, hard questionPosted: Thu, 10 May 2012 19:33:44 UTC
 S.O.S. Newbie

Joined: Thu, 10 May 2012 10:10:29 UTC
Posts: 1
According to your post the starting temperature is -32C (MINUS 32 C) and after 2 hours 19.2C (PLUS 19.2C). The rest of the temperatures are also positive. Either change the stating temperature to a 32C or the rest of the temperatures to negative ones.

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