First you need to figure out the equation for D(t)

dT/dt is proportional to (T - Ta)

Because the coffee is cooling down the sign of the derivative will be negative:

dT/dt=-k(t-Ta) from dT=-ky

You can do the integration here or you may already know that the above equation can be written in this form:

C(t)=C_{0}e^{-kt}

In this problem:

C(t) =T(t)-T_{a} = Temperature difference between coffee and temperature in the room at time t.

C_{o} = T(0) -T_{a} = T_{o}-T_{a} = Initial temperature difference at time t=0

substituting we get:

T(t)-Ta=(To-Ta)e^{-kt}

T(t)=Ta+(To-Ta)e^{-kt}

Now we need to find the k value:

dT/dt=2^{o}/min.

dT/dt=-k(t-Ta)

2^{o}/min.=-k(200-50)

k= -.0133 so now we can substitute this value for k along with the other values in our derived equation:

(t)=50+(200-50)e^{-.0133t}

(t) = 120^{o}

120^{o}=50+(200-50)e^{-.0133t}

70^{o}=150^{o}e^{-.0133t}

70/150=e^{-}^{0133t}

150/70=e^{.0133}

ln(150/70)=.0133t

t=57.3 minutes

## Comments