Question

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.)

Answer 1

Answer:

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

approximately 30 minutes.

Let $T_{0}$ = $70^{0}$F be the temperature and T that of the body

Explanation:

 Our Tm = 70, the initial-value problem is

$\frac{DT}{dt}$ = k(T − 70), T(0) = 300

Solving the equation, we get

$\frac{DT}{t-70}$ = kdt

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

    T   =  70  + $C_{2}$ $e^{kt}$

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

300=70+$C_{2}$

$C_{2}$ = 230 therefore

T= 70+ 230 $e^{kt}$

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

T (3) = 200

$e^{3k}$ = $\frac{13}{23}$

K = $\frac{1}{3}$ in $\frac{13}{23}$

= -0.19018

Therefore

T(t) = 70+230$e^{-0.19018}$