If an object at temperature \(B\) is surrounded by air at temperature \(A\), it will gradually cool so that the temperature \(T\), \(t\) minutes later, is given by Newton's law of cooling:

\[T=A+(B-A) 10^{-k t}\] The constant \(k\) depends on the particular object and can be found by solving for \(k\) \[k=\frac{1}{t} \log \frac{B-A}{T-A}\]

Use this information for Problems 34-37.

You draw a tub of hot water \((k=0.01)\) for a bath. The water is \(100^{\circ} \mathrm{F}\) when drawn and the room is \(72^{\circ} \mathrm{F}\). If you are called away to the phone, what is the temperature of the water 20 minutes later when you get in?\)