A thermometer reading 7 degrees C is brought into a room with a constant temperature of 29 degrees C.

Question

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 - (Ts-To)e^(-kt)
Ts = T surroundings = 29
To = initial Temp = 7

now put in at 4 min
T(4) = 15 = 29 - (29-7)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

Answer 1

As your cut-and-paste 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