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
May 20, 2017

Approximately 2727 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)T(t)=Ts+(T0Ts)ekt, where t_0t0 is the initial temperature of the liquid, t_sts is the temperature of the surrounding environment, tt is the number of minutes elapsed and T(t)T(t) is the temperature after tt minutes elapsed. kk is the constant, which will differ depending on the object.

We have to start by finding kk.

195 = 72 + (205 - 72)e^(-k(10))195=72+(20572)ek(10)

123/133 = e^(-10k)123133=e10k

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 180180 degrees fahrenheit, so we write the equation:

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 ~~ 26.64t26.64

Therefore, it will take approximately 2727 minutes for the coffee to cool to 180˚F.

Hopefully this helps!