Question

A solid body X of heat capacity C is kept in an atmosphere
whose temperature is T${}_A$ = 300 K. At time t = 0, the
temperature of X is T${}_0$ = 400 K. It cools according to
Newton's law of cooling. At time $t_1$, its temperature is found
to be 350 K.
At this time (t_1,) the body X is connected to a large body Y at
atmospheric temperature T${}_A$ through a conducting rod of
length L, cross-sectional area A and thermal conductivity K.
The heat capacity of Y is so large that any variation in its
temperature may be neglected. The cross-sectional area A of
the connecting rod is small compared to the surface area of X.
Find the temperature of X at time t = 3$t_1$.

Answer 1

In the first part of the question (t $\le t_1)$ At t = 0,$T_x=T_0=400Kandatt=t_1{T}_x=T_1=350K$ Temperature of atmosphere, $T_A$ - 300 K (constant) This cools down according to Newton's law of cooling. Therefore, rate of cooling temperature difference $\therefore bigg(-\frac{dT}{dt})=K(T-T_A)$ $\Rightarrow \frac{dT}{T-T_A}=-kdt$ $\Rightarrow {\int }_{T_0}^{T_1}\frac{dT}{T-T_A}=-k{\int }_0^{t_1}dt$ $\Rightarrow ln\left(\frac{T_1-T_A}{T_0-T_A}\right)=-k{t}_1$ $\Rightarrow k{T}_1=ln(2)$ .....(i) In the second part (t > t_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) $\therefore (-\frac{dT}{dt})-k(T-T_A)+\frac{KA}{CL}(T-T_A)$ ... (ii) In conduction, $\frac{dQ}{dt}=\frac{KA(T-T_A)}{L}=C(-\frac{dT}{dt})$ $\therefore (-\frac{dT}{dt})=\frac{KA}{LC}(T-T_A)$ where, C = heat capacity of body X $(-\frac{dT}{dt})=(k+\frac{KA}{CL})(T-T_A)$ ... (iii) Let at t = 3t${}_i$, temperature of X becomes T${}_2$ Then from Eq. (iii) ${\int }_{T_1}^{T_2}\frac{dT}{T-T_A}=-(k+\frac{KA}{LC}){\int }_{t_1}^{3t_1}dt$ $ln\left(\frac{T_2-T_A}{T_1-T_A}\right)=-(k+\frac{KA}{LC})(2t_1)$ $=-(2k{t}_1+\frac{2KA}{LC}{t}_1)$ or ln $\left(\frac{T_2-300}{350-300}\right)=-2ln(2)-\frac{2KA{t}_1}{LC}$ kt${}_1$= ln(2)from Eq. (i) This equation gives $T_2=(300+12.5e^{\frac{-2KA{t}_1}{CL}})K$ .