Did I do this correctly?

Question

Water initially at 130° F. is left in a room of temperature 70° F to cool.
After t minutes, the temperature T of the water is given by

 

T(t) = 70 + 60e^– 0.096 t
Find the temperature of the water 10 minutes after it is left to cool. (Round to the nearest degree.)

 

The answer I came out with was 9.29 and I rounded it to 9. Did I do this correctly?

Answer 1

No , the correct answer is 92.97^o.  The water can never cool down lower than the room temperature (70^o):

 

T(t) = 70 + 60e^-0.096*t = 70 + 60*e^-0.096*10 = 70 + 60e^-0.96 = 70 + 60*0.383 = 92.97^o