Newton's law of cooling Q (1 Viewer)

[shaon0]

Need some help.

dT/dt=k(T-P)
P=temp. of surrounding area
T=temp. of object over a function of time
k is a constant

The atmospheric temp is 30. A metal object, heated to 125 placed in the air to cool and after 1 hour is found to have cooled to 80.
Whta is the temp. after 1.5hours?

Working so far:
T=30+Ae^(kt)
When t=0.
T=30+A
After one hour:-
80=30+Ae^k
50=Ae^k
But i don't know what to do after and i'm not too sure how to approach this problem.

 

[beentherdunthat]

hint -

' heated to 125 placed in the air to cool and after 1 hour '

edit: have you done it wrong??

 

[alez]

at t=o, T=125, so 125=30+A
so A=95
then for t=1, 50=95e^k
e^k=10/19
k=ln(10/19)
k=-0.641853886
then sub in t=1.5
T=66.27381251

 

[shaon0]

at t=o, T=125, so 125=30+A
so A=95
then for t=1, 50=95e^k
e^k=10/19
k=ln(10/19)
k=-0.641853886
then sub in t=1.5
T=66.27381251

thanks.