newton's law of cooling

[Guest]

An object cools at a rate (in degreesC/min) equal to 0.10 of the difference between its temperature and that of the surrounding air. If a room is kept at 20degreesC and the temperatrue of the object is 28degreesC, what is the temperature of the object 5 minutes later?

I use the law of cooling equation:
dT/dt = k(T-Ts)
Where Ts is the surrounding temperature. And T(t) is the temperature of the object after t minutes.
And i used this equation:
dy/dt = ky


dT/dt = k(T-20)
y(0) = T(0) - 20
y(0) = 28 - 20 = 8

so i have:
dy/dt = ky and y(0) = 8

8e^(kt)
8e^(k*5)

Now i'm assuming since the problem didn't give me another temperature at another time, then k must be .10 or something dealing with that number. I didn't really understand that part of the problem.

Taking k as .1 i get:

8e^(-.1*5) = 4.85
20+4.85 = 24.85 degreesC

is that correct?

 

[galactus]

That's what I got. I hope we both didn't mess up. Don't think so. Seems OK.

\(\displaystyle \L\\\frac{dT}{dt}=k(T-20)\)

\(\displaystyle \L\\\frac{dT}{T-20}=kdt, \;\ T(0)=28\)

\(\displaystyle \L\\ln(T-20)=kt+C_{1}\)

\(\displaystyle \L\\T=20+C_{2}e^{kt}\)

When t=0, T=28

\(\displaystyle \L\\28=20+C_{2}e^{k(0)}\)

\(\displaystyle \L\\8+C_{2}e\)

\(\displaystyle \L\\C_{2}=8e^{-1}\)

\(\displaystyle \L\\T=20+\frac{8}{e}e^{\frac{1}{2}}\approx{24.85}\)