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 ?

A pan filled with hot food cools from 94^(@) C " to " 86^(@) C in 2 mi

1.	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 ?
Answer» SOLUTION :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`