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

Answer:

a. $\frac{dT}{dt}=k(T-Tm); T(0)=190$

b. $C_{0}=122$

c. $k=-0.00259$

d. $t=153.39838$ minutos

Step-by-step 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:

$Tm=68$

$\frac{dT}{dt}=k(T-Tm); T(0)=190$

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

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

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

$T(33)=122e^{33k}+68=180\\\\e^{33k}=\frac{112}{122}\\\\33k=Ln(\frac{112}{122})\\\\k=-0.00259$

d.

$150=122e^{-0.00259t}+68\\\\Ln(\frac{150-68}{122})=-0.00259t\\\\t=153.39838$