Question

You place a cup of 210oF coffee on a table in a room that is 68oF, and 10 minutes later, it is 200oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling:A.45 minutesB.33 minutesC.1 hourD.15 minutes

Answer 1

We know that Newton's law of cooling is:

`\begin{gathered} T(t)=T_A+(T_O-T_A)e^(-kt) \\ \text{ where} \\ T_A\text{ is the temperature of the surrounding} \\ T_O\text{ is the temperature of the object} \\ k\text{ is the cooling constant } \end{gathered}`

From the problem we know that the cup was originally at 210°F and that the room is at 68°F, then we have:

`\begin{gathered} T(t)=68+(210-68)e^(-kt) \\ T(t)=68+142e^(-kt) \end{gathered}`

To determine the value of k we use the fact that after 10 minutes the temperature of the object is 200°F, then we have:

`\begin{gathered} 142e^(-10k)+68=200 \\ 142e^(-10k)=200-68 \\ 142e^(-10k)=132 \\ e^(-10k)=(132)/(142) \\ -10k=\ln((66)/(71)) \\ k=-(1)/(10)\ln((66)/(71)) \end{gathered}`

Then the temperature of the cup is decreasing according to the function:

`T(t)=142e^{(1)/(10)\ln((66)/(71))t}+68`

Now, that we have the function we can determine how much time it will take for the cup to be 180°F we equate our function to this temperature and solve for t:

`\begin{gathered} 142e^{(1)/(10)\ln((66)/(71))t}+68=180 \\ 142e^{(1)/(10)\ln((66)/(71))t}=180-68 \\ e^{(1)/(10)\ln((66)/(71))t}=(112)/(142) \\ (1)/(10)\ln((66)/(71))t=\ln((56)/(71)) \\ t=(10\ln((56)/(71)))/(\ln((66)/(71))) \\ t\approx32.5 \end{gathered}`

Therefore, it will take approximately 33 minutes for the cup to be 180°F