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A body cools in a surrounding of consta...

A body cools in a surrounding of constant temperature `30^(@)C`. Its heat capacity is `2J//^(@)C`. Initial temperature of the body is `40^(@)C`. Assume Newton's law of cooling is valid. The body cools to `36^(@)C` in `10` minutes.
In further `10` minutes it will cool from `36^(@)C` to :

A

`34.8 ^(@)C`

B

`32.1 ^(@)C`

C

`32.8^(@)C`

D

`33.6^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
D
`(40-36)/(10) = k(38-30)` and `(36-x)/(10) = "k"((36+x)/(2) - 30) implies "x" = 33.6`
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