At 2 pm, a thermometer reading 80°f is taken outside where the air temperature is 20°f. at 2: 03 pm, the temperature reading yielded by the thermometer is 42°f. later, the thermometer is brought inside where the air is at 80°f. at 2: 10 pm, the reading is 71°f. when was the thermometer brought indoors?

Use Law of Cooling:

T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42


Substituting k back into cooling equation gives:


At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:

Solve for x:

Sub back into original cooling equation, x = T(t)

Solve for t:


This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

The correct answer is d. (0,5)

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