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

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

Step-by-step explanation:

Our Tm = 70, the initial-value problem is

`(DT)/(dt)` = k(T − 70), T(0) = 300

Solving the equation, we get

`(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)` =
`(13)/(23)`

K =
`(1)/(3)` in
`(13)/(23)`

= -0.19018

Therefore

T(t) = 70+230
`e^(-0.19018)`