Answer:
Here we will use the relation:
Ln(e^x) = x
The Newton's law of cooling can be written as:
T(t) = Te + (T(0min) - Te)*e^(-t/k)
Where t is time in minutes, Te is the environment temperature, T(0min) is the initial temperature and k is a constant.
In this case, we have that:
Te = 60°F
T(0min) = 150°F
now we can replace those in our equation and get:
T(t) = 60°F + (150°F - 60°F)*e^(-t/k)
T(t) = 60°F + 90°F*e^(-t/k)
Now, we know that after 15 minutes the temperature of the tea is 100°F.
T(15min) = 60°F + 90°F*e^(-15min/k) = 100°F
Whit this we can find the value of k
60°F + 90°F*e^(-15min/k) = 100°F
90°F*e^(-15min/k) = 100°F - 60°F = 40°F
e^(-15min/k) = 40°F/90°F = (4/9)
Now we can apply the Ln(x) in both sides, to get:
Ln(e^(-15min/k) ) = Ln(4/9)
-15min/k = Ln(4/9)
k = -15min/Ln(4/9) = 18.497 min.
Now we can replace that in our equation and we get:
T(t) = 60°F + (90°F)*e^(-t/18.497 min)
Now we want to find the time x such that the temperature of the tea is 80°F.
T(x) = 80°F = 60°F + (90°F)*e^(-x/18.497 min)
Now let's solve this for x.
80°F = 60°F + (90°F)*e^(-x/18.497 min)
80°F - 60°F = (90°F)*e^(-x/18.497 min)
20°F/90°F = e^(-x/18.497 min)
Now we apply Ln(x) in both sides:
Ln(2/9) = Ln(e^(-x/18.497 min)) = -x/18.497 min
ln(2/9)*18.497min = -x = -27.82 min.
x = 27.82 mins.
So 27.82 minutes after you pour the cup of tea, will be at 80°F.
But the question is:
"How long before the tea is at 80 degrees * F"
And this is asked 15 minutes after the tea was poured, so at this point the time left will be:
27.82min -15min = 12.82 min
