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The temperature of a body falls from 40^...

The temperature of a body falls from `40^(@)C` to `36^(@)C` in `5` minutes when placed in a surrounding of constant temperature `16^(@)C`. Find the time taken for the temperature of the body to become `32^(@)C`.

Text Solution

Verified by Experts

Using average form of Newton's lawof cooling, we have
`(40-36)/(5) = k((40+36)/(2)-16)` ….(1)
and `(36 - 32)/(t) = k((36 + 32)/(2) - 16)` …(2)
`((1))/((2))` gives
`(t)/(5) = (22)/(18)`
`rArr t=6.11 min`
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Knowledge Check

  • The temperature of a body falls from 50^(@)C to 40^(@)C in 10 minutes. If the temperature of the surroundings is 20^(@)C Then temperature of the body after another 10 minutes will be

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  • A body cools from 50^(@)C to 46^(@)C in 5 minutes and to 40^(@)C in the next 10 minutes. The surrounding temperature is :

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