2. Answer the following questions for the system in Fig. 2. (a) What is the common final temperature Tf? Assume the heat capacity is not a function of the temperature or volume. (Give Tf in terms of Th and Tc.) (b) Calculate the change in the entropy of the system ASsys. d) e) Th Tc v/2 v/2 v/2 V/2 v Figure 2: Now consider the mixing of two samples of the same ideal gas at two different temperatures Th and Tc, shown in c), d), and e). Initially, one mole of the gas at Th is enclosed in the left volume of , while one mole of the gas at Te is enclosed in the right side, also with volume z. Again, the system is separated from the surroundings by rigid adiabatic walls. The hatch is opened and system is allowed to mix. The hatch is then closed. (Note: the large and small circles indicate the SAME GAS at the two different temperatures initially.) Goal: (a) Find Ty (b) Find ASsys Recall: (a) When walls are adiabatic, Q = Qn+Qc = 0 = Qc = -Qh. (b) In general, the heat can be expressed as dQ = CdT, where C is the heat capacity. (c) ASsys = ASn + AS