A cup of tea cools from `80^(@)C" to "60^(@)C` in 40 seconds. The ambient temperature is `30^(@)C`. In cooling from `60^(@)C" to "50^(@)C`, it will take time:

Options

A. `50`s
B. `90`s
C. `60` s
D. `48` s

Answer

D

Explanation

Newton's law of cooling `(Deltatheta)/(Deltat)= k[0-theta_(0)]`
`theta_(0)` = surrounding's temperature implies `(80-60)/(t) = k[(80+60)/(2) - 30`] …. (i)
and `(60-50)/(t) = k[(60+50)/(2)-30`] … (ii)
`implies t = 48` sec