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

A, B, C

Explanation

Case(i)
According to Newton's law of cooling
`(dT)/(dt)=-K'(T-T_A) rArr (dT)/(T-T_A)=-K'dt`
On integrating, we get
`int_400^300 (dT)/(T-T_A)=Kint_0^(t_1)dt`
`-[log_e(T-T_A)]_400^300=K'[t]_0^(t_1)`
`rArr-loge(350-300)/(400-300)=K't_1`
`rArrlog_e100/50=K't_1 or K't_1=log_e2.....(i)`
Case (ii)
`:. -(dT)/(dt)=K'(T-T_A)+(KA(T-T_A))/(CL)`
`rArr-(dT)/(dt)=[K'+(KA)/(CL)](T-T_A) (for tgtt_1)`
Where K= coefficient of thermal conductivity of the rod.
`-(dT)/(dt)=[K'+(KA)/(CL)]dt`
On integrating, we get
`-int_350^T(dT)/(T-T_A)=int_(t_1)^(3t_1) (K'+(KA)/(CL))dt`
`rArr-[log_e(T-T_A)]_350^T=(K'+(KA)/(CL))[t]_(t_1)^(3t_1)`
`rArr(log_e) (350-300)/(T-300)=(K'+(KA)/(CL))2t=2K't_1`
`rArr(log_e)50/(T-300)=2(log_e2)+(2KA)/(CL)t_1`
`50/(T-300)=e^{log_e4+(2KA)/(CL)t_1}`
`rArrT-300=50e^-[log_e4]xxe^((-2KAt_1)/(CL))`
`rArrT=[300+12.5e^((-2KAt_1)/(CL))] Kelvin`