Hi, I have an elastic penduluum (light spring with a calorific capacity Cl, with a mass m attached on his lower end with a capacity C) attached in a cylindrical container on the upper side (a vertical elastic penduluum) , vacuum is in this cylinder .

We have to compute the difference of his internal energy (between initial state where the spring was not deformed L0 and any other length L).

second we have to compute the final temperature Tf where the spring is in equilibrium. all constants Cl , C , k, L0 are independent of temperature.

thanks

Really I have the solution but I have not understood it.

For the first question there is no problem: from the first principle the author wrote the first principle dU= delta Q + delta W delta W = k (L-L0) dL (reversible work). and delta Q = (C+Cl) dT so he found U-U0 = (C+Cl) (T-T0) + 1/2 k (L-L0)^2. with delta Q = T dS (Here is the problem the spring with mass attached on it is in vacuum so there is no heat transfer?!!).

in the second question he said that the system is themically Isolated (no transfer of heat so delta Q=0!!!! with what he wrote before why?.

We have to compute the difference of his internal energy (between initial state where the spring was not deformed L0 and any other length L).

second we have to compute the final temperature Tf where the spring is in equilibrium. all constants Cl , C , k, L0 are independent of temperature.

thanks

Really I have the solution but I have not understood it.

For the first question there is no problem: from the first principle the author wrote the first principle dU= delta Q + delta W delta W = k (L-L0) dL (reversible work). and delta Q = (C+Cl) dT so he found U-U0 = (C+Cl) (T-T0) + 1/2 k (L-L0)^2. with delta Q = T dS (Here is the problem the spring with mass attached on it is in vacuum so there is no heat transfer?!!).

in the second question he said that the system is themically Isolated (no transfer of heat so delta Q=0!!!! with what he wrote before why?.

Last edited: