Q3. When the rod is removed from contact with reservoirs it thermalised in the adiabatic environment, and its final temperature, T final, can be obtained from the conservation law of its internal energy U initial.=U final.. written in the form of.

Q4 Thus, the final rods temperature is equal to.

Q5. Consider a rod length dx at distance x from the origin, i.e. from the rod's cold end. From the general relation between heat capacity, C, and entropy, S,

Q6. one can calculate the entropies alteration for this element of the rod, viz.,

Q7. By integrating this expression over all rod's length, one can get the total alteration of the rod's entropy

Please answer each question by only giving the final answer for each of the questions. working out is not needed as only the right answer is required. Thanks, Will upvote if answers are right.

Part I the thermal contact with the reservoirs to an adiabatic environment. Calculate the alteration of the rod's entropy within its thermalisation. You may use the following parameters of the rod: Length ; Cross-sectional area A; Material density ; Specific heat c; Thermal conductance Solution: Introduce a coordinate axis, x, along the rod with an origin at the point of contact with the cold reservoir Tc. The initial temperature distribution along the rod can be obtained as a solution to the heat conduction equation jjjjjj=T=T=T=T=T=T where j is the heat flux along the rod. By applying the boundary conditions T(x=0)=Tc and T(x=)=Th to the solution, one can show that the distribution function of temperature along the rod is given by