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. 50 s
B. 90 s
C. 60 s
D. 20 s

Answer

A

Explanation

Rate of cooling`prop`excess of temperature
`(80^(@)-60^(@))/(60^(@))=K[((80^(@)+60^(@))/(2))-30^(@)]`
`(1)/(3)=K(40^(@))` . . (i)
and `(60^(@)-50^(@))/(t)=K[(60^(@)+50^(@))/(2)-30^(@)]`
`(10)/(t)=K(25)` . . (ii)
From Eqs. (i) and (ii)
`(t)/(30)=(40)/(25)impliest=(40)/(25)xx30`
`=48s=50s`