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A body cools in a surrounding of consta...

A body cools in a surrounding of constant temperature `30^(@)C`. Its heat capacity is `2J//^(@)C`. Initial temperature of the body is `40^(@)C`. Assume Newton's law of cooling is valid. The body cools to `36^(@)C` in `10` minutes.
When the body temperature has reached `36^(@)C`. it is heated again so that it reaches to `40^(@)C` in `10` minutes. Assume that the rate of loss of heat at `38^(@)C` is the average rate of loss for the given time . The total heat required from a heater by the body is :

A

`7.2J`

B

`0.728 J`

C

`16J`

D

`32J`

Text Solution

Verified by Experts

The correct Answer is:
C

`because (40-36)/(10) = k(38-30) implies k = (4)/(10xx8) = (1)/(20)`
when the block is at `38^(@)C` and temperature is at `30^(@)C` the rate of heat loss ms`xx(d theta)/(dt) = "ms" k (38-30)`
Total heat loss in `10` minutes `implies DeltaQ = "ms:k(38-30) xx 10 = 2 xx(1)/(20) xx 8 xx 10 = 8J`
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ALLEN -GEOMETRICAL OPTICS-SOME WORKED OUT EXAMPLES
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  2. A body cools in a surrounding of constant temperature 30^@C Its heat c...

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  3. A body cools in a surrounding of constant temperature 30^(@)C. Its he...

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  8. A body cools from 50^(@)C to 49.9^(@)C in 5 sec. it cools from 40^(@)C...

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  12. A container having base area A(0). Contains mercury upto a height l(0)...

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  14. One mole of an ideal monatomic gas undergoes the process P=alphaT^(1//...

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