# 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

#### 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))#

#123/133 = e^(-10k)#

#ln(123/133) = -10k#

#k = -1/10ln(123/133)#

So our formula is

#T(t) = 72 + 133e^(1/10ln(123/133)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)#

#108/133 = e^(1/10ln(123/133)t)#

#ln(108/133) = 1/10ln(123/133)t#

A good approximation gives

#t ~~ 26.64#

Therefore, it will take approximately

Hopefully this helps!