Question
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?
Answers
drwls
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*(T-T'), 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
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*(T-T'), 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