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A body takes 5 minutes to cool from 90^(...

A body takes 5 minutes to cool from `90^(@)C` to `60^(@)C` . If the temperature of the surroundings is `20^(@)C` , the time taken by it to cool from `60^(@)C` to `30^(@)C` will be.

A

5 min

B

8 min

C

11 min

D

12 min

Text Solution

Verified by Experts

The correct Answer is:
C

Since, `(T_1-T_2)/(t)=K[(T_1+T_2)/(2)-T_0]`
`(90-60)/(5)=K((90+60)/(2)-20) ...(i)
and `(60 -30)/(t) = K ((60+30)/(2)-20)` … (ii)
By (i) and (ii) -
`t/5 = (55)/(25) implies` t= 11 minutes.
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