Water initially at 190 degree F is left in a room of tempera

Water initially at 190 degree F. is left in a room of temperature 70 degree F to cool. After t minutes, the temperature T of the water is given by T(t) = 70 + 120e^-0.096 t Find the temperature of the water 10 minutes after it is left to cool. (Round to the nearest degree.) Answer: __________

Solution

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