Question

(1 point) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings.Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. Let k>0k>0 be the constant of proportionality. Assume the coffee has a temperature of 190 degrees Fahrenheit when freshly poured, and 33 minutes later has cooled to 180 degrees in a room at 68 degrees.(a) Write an initial value problem for the temperature T of the coffee, in Fahrenheit, at time t in minutes. Your answer will contain the uknown constant k :dTdt=Equation EditorT(0)=Equation Editor(b) Solve the initial value problem in part (a). Your answer will contain the unknown constant k .T(t)=Equation Editor(c) Determine the value of the constant kk=Equation Editorminutes.(d) Determine when the coffee reaches a temperature of 150 degrees.Equation Editorminutes.

Answer 1

a.
`(dT)/(dt)=k(T-Tm); T(0)=190`

b.
`C_(0)=122`

c.
`k=-0.00259`

d.
`t=153.39838\\` minutos

Explanation:

a. Newton's law of cooling states that the speed with which a body is cooled is proportional to the difference between its temperature and that of the medium in which it is found. Then, the initial value problem is given by:

`Tm=68`

`(dT)/(dt)=k(T-Tm); T(0)=190`

b. The differential equation obtained is a differential equation of separable variables:

`(dT)/(T-Tm)=kdt\\\\\int {(dT)/(T-Tm)}=\int{kdt}\\\\Ln|T-Tm|=kt+C\\\\T(t)=C_(0)e^(kt)+Tm=C_(0)e^(kt)+68\\\\T(0)=C_(0)e^(k(0))+68=190\\\\C_(0)=122`

c. After 33 minutes of serving the coffee has cooled to 180°:

`T(33)=122e^(33k)+68=180\\\\e^(33k)=(112)/(122)\\\\33k=Ln((112)/(122))\\\\k=-0.00259`

d.

`150=122e^(-0.00259t)+68\\\\Ln((150-68)/(122))=-0.00259t\\\\t=153.39838\\\\`