Question

If you pour a cup of coffee that is 200 F, and set it on desk in a room that is 68 F, and 10 minutes later it is 145 F, what temperature will it be 15 minutes after you originally poured it?

Answer 1

if you lost 55 degrees in 10 min you would loose 5.5 degrees in 1 min then you multiply by 15. 15* 5.5 = 82.5 degrees then you subtract from 200
200-82.5 = 117.5 degrees in 15 min <--------Answer

Answer 2

Newton's Law of Cooling:\[\Large\rm T(t)=T_a+(T_o-T_a)e^{-kt}\]Where:
\(\Large\rm T_a\) is the surrounding or ambient temperature and
\(\Large\rm T_o\) is the initial temperature of the thing in question (coffee in this case).

For our problem we have:\[\Large\rm T_a=68\]\[\Large\rm T_o=200\]
So this gives us an equation for temperature of:\[\Large\rm T(t)=68+(200-68)e^{-kt}\]\[\Large\rm T(t)=68+132e^{-kt}\]We're still missing our k coefficient though.

Answer 3

They tell us that at time t=10 minutes, the temperature T=145 degrees.
\[\Large\rm T(10)=145\]We'll use this to solve for our unknown k value.

Answer 4

\[\Large\rm 145=68+132e^{-10k}\]So I plugged in 10 for t,
and 145 for our T(t).

Answer 5

Do you understand how to find your k value from here?

Answer 6

k is .068