Question

A cup of coffee at 90 degrees Celsius is put into a 20 degree Celsius room when t=0 . The coffee’s temperature is changing at a rate of r(t)=-8(0.2^t) degree Celsius per minute, with t in minutes. Estimate the coffee’s temperature when t=10 . please helppp!medals are awarded!

Answer 1

integrate with respect to t i think; since you know the rate of change perhaps?

Answer 2

looks like a pain :)

Answer 3

\[-8\int (.2^t)dt\]

Answer 4

I'm not sure how to go about the limits of the integral though and I don't have the rate of change:x

Answer 5

.2^t = y

t = log.2(y)

t = ln(y)/ln(.2)

i c ant recall it too clearly either

Answer 6

r(t)=-8(0.2^t) degree Celsius per minute, with t in minutes. Estimate the coffee’s temperature when t=10

r(t) = -8(.2^(10)) = -8.192 x10^-7

Answer 7

http://www.wolframalpha.com/input/?i=-8%280.2^x%29

Answer 8

so when t = 0; we got 90 degrees.

90 = 4.97068 * .2^t +C

Answer 9

.2^0 = 1 :)

C = 90 - 4.97068 = 85.02932

h(t) = 4.97068(2^10) + 85.02932 then

Answer 10

i get: 85.0293.... for the temperature at 10 sec

Answer 11

err.. 10 minutes

Answer 12

you got that as the final? what were all the numerous steps in the link you sent me?

oh my goodness, i appreciate this so much!

Answer 13

the steps showed how they would have integrated it up to the indefinite; then you gotta pply your initial conditions to colve for the constant

Answer 14

okay, I'm still looking it over now. I understand it for the most part. Thank you so so so so much! I am truly grateful for your help:)