A pan filled with hot food cools from `94^(@)C` to `86^(@)C` in 2 minutes when the room temperature is at `20^(@)C`. How long will it take to cool from `71^(@)C` to ` 69^(@)C`? Here cooling takes place according to Newton's law of cooling.

Explanation

The average temperature of `9^(@)C" and "86^(@)C "is" 90^(@)C", which is "70^(@)C` above the room temperature. Under these conditions the pan cools `8^(@)C` in 2 minutes
Using Eq. (11.17), we have
`("Change in temperature")/("Time")KDeltaT`
`(8^(@)C)/("2 min")=K(70^(@)C)`
The average of `69^(@)C" and "71^(@)C" is "70^(@)C`, which is `50^(@)C` above room temperature. K is the same for this situation as for the original
`(2^(@)C)/("Time")=K(50^(@)C)`
When we divide above two equations, we have
`(8^(@)C//2min)/(2^(@)C//time)=(K(70^(@)C))/(K(70^(@)C))`
Time = 0.7 min
42s