Posted by Erica on Monday, April 25, 2011 at 7:54am.
The rate that an object cools is directly proportional to the difference between its temperature ( in Kelvins) at that time and the surrounding temperature (in Kelvins). If an object is initially at 35K, and the surrounding temperature remains constant at 10K, it takes 5 minutes for the object to cool to 25K. How long will it take for the object to cool to 20K?

calc  drwls, Monday, April 25, 2011 at 8:14am
You are assuming Newton's law of cooling. This leads to a differential equation of T(t) that has a simple exponential decay law. I will skip these steps.
Let k be the cooling rate constant and T' be the surrounding temperature (10 K in this case). t is time in minutes.
The temperature decay equation is
T  T' = 25 e^(kt)
where the 25 K is the initial temperature difference relative to the surroundings.
Note that dT/dt = k*(TT'), as required by the cooling law, and that T(t=0) = 25 + T'
15 = 25*e^(5k)
5k = ln(0.6)
k= 0.102 min^1
When T = 20K,
10 = 25*e^(0.102 t)
0.4 = e^(0.102 t)
.102t = ln(0.4)
t = 9.0 minutes
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