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Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionality, T(t) is the temperature of the object for t  0, and Tm is the ambient temperature –that is, the temperature of the medium around the object. When a cake is removed from an oven, its temperature is measured at 300 . 0 F Three minutes later its temperature is F 0 200 . How long will it take for the cake to cool off to a room temperature of F 0 70 ? (Assume that Tm =70.)

Respuesta :

richard5salguero

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let [tex]T_{0}[/tex] = [tex]70^{0}[/tex]F be the temperature and T that of the body

Explanation:

 Our Tm = 70, the initial-value problem is

[tex]\frac{DT}{dt}[/tex] = k(T − 70), T(0) = 300

Solving the equation, we get

[tex]\frac{DT}{t-70}[/tex] = kdt

In [T-70]= kt +[tex]C_{1}[/tex]

    T   =  70  + [tex]C_{2}[/tex] [tex]e^{kt}[/tex]

Finding he value for [tex]C_{2}[/tex] using the initial value of T (0)= 300, therefore we get:

300=70+[tex]C_{2}[/tex]

[tex]C_{2}[/tex] = 230 therefore

T= 70+ 230 [tex]e^{kt}[/tex]

Finding the value for k using T (3)  = 200, therefore we get

T (3) = 200

[tex]e^{3k}[/tex] = [tex]\frac{13}{23}[/tex]

K = [tex]\frac{1}{3}[/tex] in [tex]\frac{13}{23}[/tex]

= -0.19018

Therefore

T(t) = 70+230[tex]e^{-0.19018}[/tex]

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