Question

a pizza is removed at 1:00 PM from an oven whose temperature is fixed at 425 degrees f into a room that is a constant 71 degrees. after 5 minutes, the pizza pan is at 300 degrees. a). at what time is the temperature of the pan 135 degrees

Answer 1

11.18 minutes

Step-by-step explanation:

We shall apply Newton's law of cooling

Rate of cooling ∝ Temperature difference

(T₂ - T₁ ) / t = k ( T₁ + T₂ / 2 - T )

A body cools from temperature of T₂ TO T₁ in time t , K being a constant.

(T₁ + T₂) /2 is the average temperature and T is temperature of surrounding.

Using the data given in the problem ,

For cooling from 425 degree to 300 degree

`(425-300)/(5) =k((425+300)/(2)-71)`

`(125)/(5)= k x 291.5`

For cooling from 425 to 135 in time t we have

`(425 - 135)/(t) =k((425+135)/(2)-71)`[/tex]

`(290)/(t)` = 209k

From these two equations we get the value of t as

t = 16.18 s

So time required to cool from 300 to 135 degree isas follows

16.18 - 5 = 11.18 s