# Newton's law of cooling question/problem

Printable View

• July 18th 2010, 11:28 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

Please help me
• July 19th 2010, 04:40 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

Please help me

First, your formula is wrong. At least one of the "t" should be "T", the temperature of the environment, 70 degrees, while at least one should be "t" the time after the object is removed from cooling.

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= 0, when the object is first taken out of cooling, U(0)= 28 so [tex]28= 70+ (U0- 70)e^{k(0)}= 70+ (U0- 70)= U0. That tells you what U0 is.

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 $U(30)= 70+ (U0- 70)e^{30k}$