A cup of tea cools from `80^(@)C` to `60^(@)C` in one minute. The ambient temperature is `30^(@)C` . In cooling from `60^(@)C` to `50^(@)C` it will take

Options

A. 30 seconds
B. 60 seconds
C. 96 seconds
D. 48 seconds

Answer

D

Explanation

`(80-60)/(1)=K((80+60)/(2)-30)` …(i)
`(60-50)/(t)=K((60+50)/(2)-30)` …(ii)
Solving (i) and (ii)
`t=4/(5)`minute = 48 second