InterviewSolution
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InterviewSolution
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The rate at which a substance cools in moving air is proportional tothe difference between the temperatures of the substance and that of the air.If the temperature of the air is 290 K and the substance cools from 370 K to330 K in 10 min, when will the temperature be 295 K? |
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Answer» Let T be the temperature of the substance at time t. `therefore -(dT)/(dt) = k(T-290)` or `(dT)/(dt) =-(T-290)` (Negative sign because `(dT)/(dt)` is rate of cooling) or `int(dT)/(T-290)=-kint(dt)` Integrating the L.H.S between the limits `T=370` to T=330 and the R.H.S between the limits t=0 to t=10, we get `int_(370)^(330) (dT)/(t-290)=-k_(int_(0)^(10))dt` or `log40-log80 =-k xx 10` `log2=10k` or `k=(log2)/10`..............(2) Now, integrating equation (1) between `T=370` and `T=295` and t=0 and t=t, we get `int_(370)^(295)(dT)/(T-290)=-kint_(0)^(1)dt` or `log(T-290)|underset(370)overset(295)(=-kt` or `t=(log 16)/k` Hence, from equation (2), we get `t=(log16)/k` Hence, from equation (2), we get `t=(log 16)/(log2) xx 10 = 40` i.e., after 40 min |
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