Differential Equations A cup of hot chocolate is initially a

A cup of hot chocolate is initially at 110 degree F and is left in a room with ambient temperature of 70 degree F. Suppose that after one minute, the chocolate cooled 10 degree F. Write down the differential equation satisfied by the temperature of the chocolate. Then solve it. How long does it take the hot chocolate to cool to 75 degree F?

Solution

By Newton’s law of cooling :The rate of cooling is proportional to the the difference between the current temperature and the ambient temperature.

We first Write an initial value problem that models the temperature of the hot chocolate.

LetT(t) be the temperature of the hot chocolate at time t,and A be the ambient temperature of the room,then

dT/dt = -k(T-A). where k is the proportionality constant.

Initially the chocolate is cooling at the rate of 10 degrees per minute

Hence,

-10= -k(T(0) - 70)

-10 = -k(110 - 70)

k = 0.25

Our differential equation is

dT/dt = -0.25(T-70)

We solve the above equation by method of separation of variables to get

T(t) = ce^-0.25t + 70 where c is the integration constant

Substituting the initial condition T(0)=110 we get

110=c+70

c=40

T(t) = 40e^-0.25t + 70 is our solution to the differential equation.

We must solve

75 = 40e^-0.25t + 70

5= 40e^-0.25t

0.125= e^-0.25t

ln(0.125)=-0.25t

t = 8.32 minutes

It takes approximately 8 minutes 19 seconds for the hot chocolate to cool down to 75