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

117.5F

Answer 2

how did you get it?

Answer 3

@caesar27

Answer 4

i get 127

Answer 5

of course i could be wrong

Answer 6

@satellite73 can you explain how you got it please? :)

Answer 7

1 minute -5.5F
so 10 minute = 145F so that ( -5.5 x 10 minute=-55F)
so 15 minute = 117.5F so that ( -5.5 x 15 minute =-82.5F)

Answer 8

sure but it will take a bit of explanation
first off, you work always with the difference in the temp of the heated object, and the room temperature. it is the difference in the temperature that decays (goes to zero)

Answer 9

@caesar27 it is not linear

Answer 10

of course I could be wrong, please you practice

Answer 11

the initial temperature difference is \(200-68=132\) and this is the part that decreases to zero
after 10 minutes it is 145 degrees and \(145-68=77\) so the snap way to do this is say the model for the difference in the temperature is
\[132\left(\frac{77}{132}\right)^{\frac{t}{10}}\]

Answer 12

you want the temp at \(t=15)\) so compute
\[132\left(\frac{77}{132}\right)^{\frac{15}{10}}\]

Answer 13

then since this is the difference in the temperatures, add 68 to get your result
http://www.wolframalpha.com/input/?i=132%2877%2F132%29^%283%2F2%29%2B68

Answer 14

this is probably not how you were taught to do it. you probably had to come up with an equation that looks like
\[132e^{kt}\] but that requires some work to find \(k\) using logs

Answer 15

the way you showed me should be fine, thank you so much.. can you possibly help me with another problem i will post? @satellite73

Answer 16

sure