My hot venti Starbucks coffee measured 70 °C when I decided to make it an iced c

Question

My hot venti Starbucks coffee measured 70 °C when I decided to make it an iced coffee. So put the 70 °C coffee in a refrigerator with a temperature of 5 °C and after 30 minutes the coffee cooled to a temperature of 14 °C.

a. Use Newton's cooling law to write a first order differential equation for the change in the temperature of the coffee T as a function of time t.

b. Use the information given to solve the differential equation for T. c. What was the temperature of the coffee after 60 minutes?

Explanation / Answer

Newtons law of cooling dT/dt = -k(T-T0) where T0 is the surrounding temperature =5 deg ,T is the instantaneous Temperature and 'k' is positive constant. On integration, we get,

=> -kt = log(T-T0)+C

=>T-T0 =C*e^(-kt) , where C is the integration constant.

using the initial condition, @t=0 , T=T1 =70 deg

=> C=T1-T0

=> T= T0+(T1-T0)e^(-kt)

and putting the values

T= 5+(70-5)e^(-kt) =5+65e^(-kt)

b. It is given that @t=30 min T= 14 deg

Putting in the above equation, we get

14=5+65e^(-k30)

=> k =0.0659/min

@t= 60min

T= 5+65*e^(-0.0659*60) =6.2465 deg

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