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. 30 s

Answer

A

Explanation

According to Newton 's law of cooling ,
Rate of cooling `prop` excess of temperature
Case I `(80-60)/(60)=k[((80+60)/(2))-30^(@)C]`
`(1)/(3)=k(40)`
Case II `(60-50)/(t)=K((60+50)/(2)-30^(@)C)`
`(10)/(t)=Kxx25` …(ii)
Dividing Eq . (i) by Eq. (ii), we get
`(t)/(30)=(40)/(25)rArrt=(40)/(25)xx30=48~~50s`