A solid body X of heat capacity C is kept in an atmosphere whose temperature is `T_A=300K`. At time `t=0` the temperature of X is `T_0=400K`. It cools according to Newton's law of cooling. At time `t_1`, its temperature is found to be 350K. At this time `(t_1)`, the body X is connected to a large box Y at atmospheric temperature is `T_4`, through a conducting rod of length L, cross-sectional area A and thermal conductivity K. The heat capacity Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of hte connecting rod is small compared to the surface area of X. Find the temperature of X at time `t=3t_1.`

Answer

`(300+12.5e"^((-2KAt_(1)))/(CL))`

Explanation

`(300+12.5_(e)^((-2KAt_(1))/(CL)))`
In the first part of the question `(tlet_(1))`
`(T_(A)=300K)/(X)`
At `t=0,T_(X)=T_(0)=400Kandat t=t_(1),T_(X)=T_(1)=350K`
Temperature of atmosphere, `T_(A)=300K` (constant)
This cools down according to Newton’s law of cooling.
Therefore, rate of cooling `prop` temperature difference.
`therefore(-(dT)/(dt))=k(T-T_(A))rArr(dT)/(T-T_(A))=kdtrArrint_(T_(0))^(T_(1))(dT)/(T-T_(A))=-kint_(0)^(t_(1))dt`
`rArrln((T_(1)-T_(A))/(T_(0)-T_(A)))=-kt_(1)rArrkt_(1)=-ln((350-300)/(400-300))`
`rArrkt_(1)=ln(2)` ... (i)
In the second part `(gtt_(1))` , body X cools by radiation (according to Newton’s law) as well as by conduction.
Therefore, rate of cooling = (cooling by radiation) + (cooling by conduction)

In conduction `,(dQ)/(dt)=(KA(T-T_(A)))/(L)=C(-(dT)/(dt))therefore(-(dT)/(dt))=(KA)/(LC)(T-T_(A))`
where, C = heat capacity of body X
`therefore(-(dT)/(dt))=k(T-T_(A))+(KA)/(CL)(T-T_(A))`... (ii)
`(-(dT)/(dt))=(k+(KA)/(CL))(T-T_(A))`...(iii)
Let at `t=3t_(1)` , temperature of x becomes `T_(2)`
Then from Eq. (iii)
`int_(T_(1))^(T_(2))(dT)/(T-T_(A))=-(k+(KA)/(KC))int_(t_(1))^(3t_(1))dt`
`ln((T_(2)-T_(A))/(T_(1)-T_(A)))=-(k+(KA)/(LC))(2t_(1))=-(2kt_(1)+(2KA)/(LC)t_(1))`
`orln((T_(2)-300)/(350-300))=-2ln(2)-(2KAt_(1))/(LC),kt_(1)=ln(2)` from Eq . (i) .
This equation gives `T_(2)=(300+12.5e^((-2KAt_(1))/(CL)))K`