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A body cools from 50^(@)C to 46^(@)C in ...

A body cools from `50^(@)C` to `46^(@)C` in 5 min and to `40^(@)C` in the next 10 min. then the temperature of the surrounding is

A

`30^(@)C`

B

`36^(@)C`

C

`28^(@)C`

D

`32^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
`((d theta)/(dt))_(1)=k(theta_(AV)-theta_(0))`
`(4)/(5)=k(48-theta_(0))` . . . (i)
`(6)/(10)=k(43-theta_(0))` . . . (ii)
Solving equation (i) and (ii),
`theta_(0)=28^(@)C`.
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