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A body cools from 100^(@)C to 70^(@)C in...

A body cools from `100^(@)C` to `70^(@)C` in 8 second. If the room temperature is `15^(@)C`, and assuming that Newton's law of cooling holds good, then time required for the body to cool from `70^(@)C` to `40^(@)C` is

A

7 s

B

14 s

C

10 s

D

20 s

Text Solution

Verified by Experts

The correct Answer is:
B
`30/8=K[85-15]=70K`
`K=30/(8xx70)=3/56`
and`(70-40)/t=(55-15)xxK`
`therefore30/t=3/56xx40/1`
`thereforet=14s`
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MARVEL PUBLICATION-KINETIC THEORY OF GASES ,THERMODYNAMICS AND RADIATION-TEST YOUR GRASP - 9
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