Question

A pizza pan is removed at

2 : 00 PM2:00 PM

from an oven whose temperature is fixed at

425 degrees  into a room that is a constant

71 degrees   

After 5​ minutes, the pizza pan is at 300 degrees F.300°F.

​(a) At what time is the temperature of the pan 125 degrees F?

​(b) Determine the time that needs to elapse before the pan is 200 degrees .

​(c) What do you notice about the temperature as time​ passes?

Answer 1

According to Newtons law of Cooling

The temperature after t minutes is given by

In this case,

= 71 degrees

= 425 degrees

Therefore,

=>

Given, when t = 5 minutes, T(t) = 300 degrees

Therefore,

=>

=>

=>

=>

=> -5k = ln (229 / 354)

=> k = 0.087115

Therefore,

a) Let us assume the temperature is 125 degrees after n minutes

Therefore,

=>

=>

=>

=>

=> - 0.087115n = ln (54 / 354)

=> n = 21.58 minutes

Therefore, the temperature is 125 degrees after 21.58 minutes i.e. at 2:22 PM

b) Let us assume the temperature is 200 degrees after n minutes

Therefore,

=>

=>

=>

=>

=> - 0.087115n = ln (129 / 354)

=> n = 11.59 minutes

Therefore, the temperature is 200 degrees after 11.59 minutes

c) We notice as the time passes, the temperature of the pizza pan decreases.