Answer :
Answer:
a) The solution of the differential equation is [tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex].
b) The reading after 20 minutes is approximately 70.770 ºF.
Explanation:
a) Newton's law of cooling is represented by the following ordinary differential equation:
[tex]\frac{dT}{dt} = -\frac{T-M}{\tau}[/tex] (Eq. 1)
Where:
[tex]\frac{dT}{dt}[/tex] - Rate of change of temperature of the object in time, measured in Fahrenheit per minute.
[tex]\tau[/tex] - Time constant, measured in minutes.
[tex]T[/tex] - Temperature of the object, measured in Fahrenheit.
[tex]M[/tex] - Medium temperature, measured in Fahrenheit.
Now we proceed to solve the differential equation:
[tex]\frac{dT}{T-M} = -\frac{t}{\tau}[/tex]
[tex]\int {\frac{dT}{T-M} } = -\frac{1}{\tau} \int \, dt[/tex]
[tex]\ln (T-M) = -\frac{t}{\tau} + C[/tex]
[tex]T(t) -M = (T_{o}-M)\cdot e^{-\frac{t}{\tau} }[/tex]
[tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)
Where:
[tex]t[/tex] -Time, measured in minutes.
[tex]T_{o}[/tex] - Initial temperature of the object, measured in Fahrenheit.
b) From (Eq. 2) we obtain the time constant of the cooling equation for the object: ([tex]M = 70\,^{\circ}F[/tex], [tex]T_{o} = 100\,^{\circ}F[/tex], [tex]t = 6\,min[/tex], [tex]T(t) = 80\,^{\circ}F[/tex])
[tex]80\,^{\circ}F = 70\,^{\circ}F + (100\,^{\circ}F-70\,^{\circ}F)\cdot e^{-\frac{6\,min}{\tau} }[/tex]
[tex]e^{-\frac{6\,min}{\tau} } = \frac{80\,^{\circ}F-70\,^{\circ}F}{100\,^{\circ}F-70\,^{\circ}F}[/tex]
[tex]e^{-\frac{6\,min}{\tau} } = \frac{1}{3}[/tex]
[tex]-\frac{6\,min}{\tau} = \ln \frac{1}{3}[/tex]
[tex]\tau = -\frac{6\,min}{\ln \frac{1}{3} }[/tex]
[tex]\tau = 5.461\,min[/tex]
The cooling equation of the object is [tex]T(t) = 70 +30\cdot e^{-\frac{t}{5.461} }[/tex] and the temperature of the object after 20 minutes is:
[tex]T(20) = 70+30\cdot e^{-\frac{20}{5.461} }[/tex]
[tex]T(20) \approx 70.770\,^{\circ}F[/tex]
The reading after 20 minutes is approximately 70.770 ºF.