If a hot cup of coffee, initially at 190◦F, cools to 125◦F in 5 minutes when places in a room with a constant temperature of 75◦F, how long will it take the coffee to reach 100◦F.

Question

I do not know how to set up the problem correctly, I have tried multiple times but always end up with the wrong answer. I know the formula but not sure about what value goes where

Answer 1

so, which formula are you using? You expect help, but do not show what you have tried so far. If using Newton's law of cooling, then you have
T(t) = 75+(190-75)e^(-t/r) = 75 + 115e^(-t/r)
You need to find r. Since T(5) = 125,
75 + 115e^(-5/r) = 125
so r = 5/log(23/10) = 6
That means T(t) = 75 + 115e^(-t/6)
and T will bee 100 when
75 + 115e^(-t/6) = 100
t = 6log(23/5) = 9.156 minutes

google can provide further explanations and examples of this formula.

Answer 2

Assuming that you will be given the formula based on Newton's Law of Cooling, which is
T(t) = c e^(-kt) + Ta, where T is temperature, t is time, and Ta is the ambiant tempereature, c and k are constants.
(you titled your post as Pre Calculus, so I assume you don't have to arrive at this formula)


given:
Ta = 75
T(0) = 190
T(5) = 125
we want T(t) = 100

T(0) = c e^0 + Ta
190 = c + 75
c = 115

so T(t) = 115 e^(-kt) + 75
when t = 5, T(5) = 125
125 = 115 e^(-5k) + 75
e(-5k) = 50/115 = .43478..
using ln and log rules
-5k = ln .43478...
k = .16658...

T(t) = 115 e^(-.16658..t) + 75

100 = 115 e^(-.16658..t) + 75
e^(-.16658..t) = 25/115 = .21739...
solving this for t , once again using natural logs, I got
t = 9.16

check my calculations

Answer 3

ahhh, I agree with oobleck, it must be correct