A liquid cools from `70^(@)C` to `60^(@)C` in `5` minutes. Calculate the time taken by the liquid to cool from `60^(@)C` to `50^(@)C` , If the temperature of the surrounding is constant at `30^(@)C` .

Answer

`7 minutes`

Explanation

`(70 -60)/(5) = K [(70 +60)/(2)-30]`
`rArr (10)/(5) = K [65 - 30] ….(i)`
Now `(60 -50)/(t) =K [65 - 30] …(ii)`
Dividing equation (i) and (ii)
`(t)/(5) = (35)/(25)`
`t = (7)/(5) xx5 = 7 min`