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A cup of tea cools from 80^(@)C to 60^(@...

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

A

`50`s

B

`90`s

C

`60` s

D

`48` s

Text Solution

Verified by Experts

The correct Answer is:
D
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
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