Question

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?

Answer 1

Approximately `27` minutes.

Explanation:

This is a Newton's Law of Cooling Problem.

Newton's Law of Cooling states that an object cools down by the formula `T(t) = T_s + (T_0 - T_s)e^(-kt)`, where `t_0` is the initial temperature of the liquid, `t_s` is the temperature of the surrounding environment, `t` is the number of minutes elapsed and `T(t)` is the temperature after `t` minutes elapsed. `k` is the constant, which will differ depending on the object.

We have to start by finding `k`.

`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` degrees fahrenheit, so we write the equation:

`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 `27` minutes for the coffee to cool to `180˚`F.

Hopefully this helps!