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Time taken by the body to cool from 45^(@)C to 35^(@)C is |
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Answer» 5 min AVERAGE temperature `=(50+40)/(2)=45^(@)` `:.(dT)/(dt)=-K(T-T_(c))` `(10)/(5)=-k(45-18)`. . . (i) `dT=45-35=10^(@)C` Average temperature, `(45+35)/(2)=40^(@)C` `:.(dT)/(dt)=-k[T-T_(c)]` `(10)/(t)=-k[40-18]`. . .(ii) DIVIDING (i) by (ii) `t/5=(45-18)/(40-18)` `rArrt=5xx(27)/(22)~=6` min So correct is (c ). |
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