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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 minBy Newton's law of coolingdθ2θ2−θ1=kdtwhere 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 minutesdθ2=61oC−59oC=2oC, θ1=30oC, θ2=61+592=60oC, dt=4min So, dθ2θ2−θ1=kdt ⇒ 260−30=k×4k=160oCminThen time taken by the body to cool it from 51oC to 49oC: dθ2θ2−θ1=kdt250−30=160×dtdt=6 min

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