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A hot liquid kept in a beaker 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)]` where `theta_(0)` is temperature of currounding. `:.(80-70)/(2)=K[(80+70)/(2)-30]` . . . (i) Also `(60-50)/(t)=K[(60+50)/(2)-30]`. . . (II) DIVIDING (i) by (ii) we have : `t/2=(75-30)/(55-30)rArr(t)/(2)=(45)/(25)` `rArrt=(45)/(25)xx2=(90)/(25)=(18)/(5)=3*6` minutes. `=216` seconds correct CHOICE is (d). |
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