Posted by **Sushmitha** on Saturday, October 29, 2011 at 9:45pm.

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 - (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

- algebra - more help please -
**Reiny**, Saturday, October 29, 2011 at 11:56pm
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

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