A bottle of white wine at room temperature (68°F) is placed in a refrigerator at 4 P.M. Its temperature after t hr is changing at the rate of −18e^(−0.6t)

room-temperature
refrigerator

#1

A bottle of white wine at room temperature (68°F) is placed in a refrigerator at 4 P.M. Its temperature after t hr is changing at the rate of **
−18e^(−0.6t)

°F/hour. By how many degrees will the temperature of the wine have dropped by 7 P.M.? (Round your answer to one decimal place.) **
____ °F

What will the temperature of the wine be at 7 P.M.? (Round your answer to one decimal place.) _____ °F

Answer:

Use newtons law of cooling… formula after derivation of dT/dt = -k(T-Ta)
will be T(t)=Ce^(-kt)

you are given the formula but in pieces, you have the rate at which the temp is changing but not the initial value for Ta which is calculated from the second bit of info gave you the starting temp or the initial ambient temp.

so… we get the cooling equation for any value “t” such that T(t)=((-18e^(-0.6t))+68))
plug in 7 into the equation for find temp after seven hours.

you get T(7)=((-18e^(-0.6*7))+68
= 67.7 deg F
so it dropped by (68-67.7). which is 0.3 deg F

to find at 7pm you just need to know the time again so find the difference from where it started cooling and when it becomes seven pm. 4pm to 7pm you get 3 hours. so find after 3 hours what the temp would be plug in 3 to the equation.

T(3)= ((-18e^(-0.6*3))+68) = 53.0 rounded to nearest tenth of degree