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`?

Explanation

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"`