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

Answer:

11.18 minutes

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

$\frac{425-300}{5} =k(\frac{425+300}{2}-71)$

$\frac{125}{5}= k x 291.5$

For cooling from 425 to 135 in time t we have

$\frac{425 - 135}{t} =k(\frac{425+135}{2}-71)$[/tex]

$\frac{290}{t}$ = 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