Asked by Anonymous
                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.
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
            
        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
Answers
                    Answered by
            oobleck
            
    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.
    
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.
                    Answered by
            mathhelper
            
    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
    
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
                    Answered by
            mathhelper
            
    ahhh, I agree with oobleck, it must be correct
    
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