# Newton's law of cooling question/problem

• Jul 18th 2010, 10:28 PM
IneeedHelp
Newton's law of cooling question/problem
An object was cooled to 28 Degrees farenheit.
It was removed from cooling and was taken into an environment of 70 Degrees farenheit.
After 10 minutes its Temperature raised to 35 Degrees farenheit.
What will its temp be after 30 Minutes

Formula: U(t)=t + (U0 - t)e^kt

• Jul 19th 2010, 03:40 AM
HallsofIvy
Quote:

Originally Posted by IneeedHelp
An object was cooled to 28 Degrees farenheit.
It was removed from cooling and was taken into an environment of 70 Degrees farenheit.
After 10 minutes its Temperature raised to 35 Degrees farenheit.
What will its temp be after 30 Minutes

Formula: U(t)=t + (U0 - t)e^kt

If I am reading this correctly, your formula should be $U(t)= T+ (U0- T)e^{kt}$ where "T" is 70 and "t" is the number of minutes after the object is removed from cooling.
At t= 10, 10 minutes after being taken out of cooling $U(10)= 70+ (U0- 70)e^{10k}= 35$. You already know U0 so you can solve that for k (you will need to use a logarithm).
Finally, after you know U0 and k, evaluate $U(30)= 70+ (U0- 70)e^{30k}$