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 `94^(@) 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. (11.21), we have
` "Change in temperature "/"Time" = K Delta T`
` (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//2 "min")/(2^(@) C//"time") = (K(70^(@) C))/(K(50^(@) C))`
Time ` = 0.7` min
` = 42 s`