A hot water cools from 92^(@)C to 84^(@)C in 3 minutes when the room temperature is 27^(@)C. How long will it take for it to cool from 65^(@)C to 60^(@)C?

A hot water cools from 92^(@)C to 84^(@)C in 3 minutes when the room t

1.	A hot water cools from 92^(@)C to 84^(@)C in 3 minutes when the room temperature is 27^(@)C. How long will it take for it to cool from 65^(@)C to 60^(@)C?
Answer» Solution :The HOT water cools `8^(@)C` in 3 minutes. The average temperature of `92^(@)C` and `84^(@)C` is `88^(@)C`. This average temperature is `61^(@)C` above room temperature. By using equation, `(dT)/(T-T_(s))=-(a)/(MS)dt or (dT)/(dt)=-(a)/(ms)(T-T_(s))` `(8^(@)C)/("3 min")=-(a)/(ms)(61^(@)C)"...(1)"` Similarly the average temperature of `65^(@)C and 60^(@)C`. The average temperature is `35.5^(@)C` above the room temperature. Then we can write `(5^(@)C)/(dt)=-(a)/(ms)(35.5^(@)C)"....(2)"` By dividing both the equation, we get `((8^(@)C)/("3 min"))/((5^(@)C)/(dt))=-((a)/(ms)(61^(@)C))/(-(a)/(ms)(35.5^(@)C))` `(8xxdt)/(3xx5)=(61)/(35.5)` `dt=(61xx15)/(35.5xx8)=(915)/(284)="3.22 min"`