You place a cup of 205°F coffee on a table in a room that is 72°F, and 10 minutes later, it is 195°F. Approximately how long will it be before the coffee is 180°F?
1 Answer
Approximately
Explanation:
This is a Newton's Law of Cooling Problem.
Newton's Law of Cooling states that an object cools down by the formula
We have to start by finding
195 = 72 + (205 - 72)e^(-k(10))195=72+(205−72)e−k(10)
123/133 = e^(-10k)123133=e−10k
ln(123/133) = -10kln(123133)=−10k
k = -1/10ln(123/133)k=−110ln(123133)
So our formula is
T(t) = 72 + 133e^(1/10ln(123/133)t)T(t)=72+133e110ln(123133)t
We're looking for the amount of time it takes for the coffee to cool to
180 = 72 + 133e^(1/10ln(123/133)t)180=72+133e110ln(123133)t
108/133 = e^(1/10ln(123/133)t)108133=e110ln(123133)t
ln(108/133) = 1/10ln(123/133)tln(108133)=110ln(123133)t
A good approximation gives
t ~~ 26.64t≈26.64
Therefore, it will take approximately
Hopefully this helps!