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A hot liquid kept in a breaker cools from 80^(@)C to 70^(@)C in two minutes. If the surrounding temperature is 30^(@)C, then the time of cooling of the same liquid from 60^(@)C to 50^(@)C is : |
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Answer» 240 s `(theta_(1)-theta_(2))/(t)=k[(theta_(1)+theta_(2))/(2)-theta_(0)]` `:.(80-70)/(2)=k[75-30]`. . . (i) `(60-50)/(t)=k[55-30]`. . .(ii) Dividing (i) by (ii) `t/2=(45)/(25)rArrt=(18)/(5)=3*6" min = 216 SECONDS"` So correct choice is (d). |
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