Posted by Sushmitha on .
A thermometer reading 7 degrees C is brought into a room with a constant temperature of 29 degrees C.
If the thermometer reads 15 degrees C after 4 minutes, what will it read in 6 minutes? 11 minutes?
So I guess I need to put into newton's law of cooling:
u(t) = T + (u sub zero  T ) e^kt
How do I solve for k so can plug in my T's. and what is u sub zero? initial what?
algebra  help please  Damon, Saturday, October 29, 2011 at 6:24pm
well I put the law like this:
T(t) = Ts  (TsTo)e^(kt)
Ts = T surroundings = 29
To = initial Temp = 7
now put in at 4 min
T(4) = 15 = 29  (297)e^4k
15 = 29  22 e^(4k)
14 = 22 e^4k
 .636363... = e^4k
ln (.636363 ... ) = 4k
4k = .452
k = .113
so you go on and put 6 minutes in for t
algebra  more help please  Sushmitha, Saturday, October 29, 2011 at 9:31pm
Still lost. I put in 6 and lost a variable.
T(6) = 29  (29  7) e^0.678
29  22e^0.678
29 = 22e^0.678
ln 1.318181818 = e^0.678
0.276253377 = 0.678

algebra  more help please 
Reiny,
As your cutandpaste shows, Damon has found the value of k to be .113
(There is a typo in his solution...
should say:
15 = 29  22 e^(4k)
14 =  22 e^4k
.636363... = e^4k
ln (.636363 ... ) = 4k
4k = .452
k = .113 )
so your equation is
T(t) = 29  22e^.113t
so t(4) = 29  22e(.113(4))
= 29  22 e^.45198
= 15