The rate at which a substance cools in moving air is proportional to the 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 to 330 K in 10 min, when will the temperature be 295 K?

Explanation

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