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Question

A body in a laboratory takes 4 minutes to cool from 61oC and 59oC. If the laboratory temperature is 30oC then the time taken by it to cool from 51oC to 49oC will be:

A
8 min
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B
6 min
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C
5 min
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D
4 min
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Solution

The correct option is A 6 min
By Newton's law of cooling
dθ2θ2θ1=kdt
where k is inversely proportional to product of mass and specific heat of the body, θ1 and θ2 are surrounding and average body temperatures and dt is the time interval.
For the first 4 minutes
dθ2=61oC59oC=2oC, θ1=30oC, θ2=61+592=60oC, dt=4min
So, dθ2θ2θ1=kdt 26030=k×4k=160oCmin
Then time taken by the body to cool it from 51oC to 49oC:
dθ2θ2θ1=kdt25030=160×dtdt=6 min

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