A hot liquid kept in a beaker cools from `80^(@)C` to `70^(@)C` in two minutes. If the surrounding temperature is `30^(@)C`, find the time of coolilng of the same liquid from `60^(@)C` to `50^(@)C`.

Options

A. 240s
B. 360s
C. 480s
D. 216s

Answer

D

Explanation

According to Newton.s law of cooling `(T_(1) -T_(2))/(t) = K ((T_(1) + T_(2))/(2) - T_(S))`
where `T_(s)` is the surrounding temperature.
For the `1^(st)` case
`(80-70)/(2) = K ((80 + 70)/(2) -30) or 5 = K (45)`...(i)
For the `2^(nd)` case
`(60-50)/(t) = K ((60 + 50)/(2)- 30) or (10)/(t) = K (25)`...(ii)
Divide (i) by (ii), we get
`(t)/(2) = (45)/(25) or t= 3.6` min= 216 s