please ahow how to find the given answers with the provided tips to approach

answer for a) is around 12 kJ

treated as a closed system, a constant value heat capacity is needed to calculate the internal energy and entropy differences between state 2 and the dead state (entropy will have 1.0115 kJ/kg-K. which can be plugged into the exergy equation with the internal energy, pressure X specefic volume, and temprature X entropy terms all present. the Po and To values will be the dead state values of 1 bar (100 kpa) and 293 K. each term is written out in terms of the three states in the problem, e.g. (u1 - u2 + u2 - uo).

also the intial internal energy, volume, and entropy should be found on a superheated steam table. a saturated liquid at 1 bar and 99.6 C are avaialble for the hypothetical state 2. also consider the process in which the dead state to state 2 will reversibly heat water feom 223 K and 1 bar to saturated liquid at 1 bar. which the changes in energy and entropy can be calculated assuming that volume change is negligble over the twmprature range. using hypothetical state 2,, the quantities in the exergy equation will be calculated as, e.g. internal energy, (U1-U2+U2-Uo = U1-U0). which the exergy should be calculated in units of kJ as the already given answer suggests.

answer for b) is around 5.9 kJ

interpolate the internal energy and entropy in the auperheated steam table to the point double the initial volume at the same pressure. Then the exergy can be calculated.

need to be using the three-term exerfy equation as in a) but only need to use the initial and final values for internal energy, specefic volume, and entropy. The final state of an isobaric heating until volume doubles can be found in the auperheated steam table by interpolating to the point of double the initial volume at the same pressure. This state is about 23% of the way from 600 C and 700 C ar 1 MPa (internal energy interpolated is 3,338 kJ/kg).

answer for c) is around -1 kJ

same as process b) but interpolate with consrant tempratureinstead of pressure

the final state of an isothermal expansion intil volume doubles can be found using the superheated steam table by interpolating to the point of double the initial volume at the same temprature. This staye is 82% of the way from 600 kpa to 500 kpa (entropy interpolated will be 7.04 kJ/kg-K)

1 mol of steam is initially at 10bar and 200C. The surroundings are at 20C and 1 bar. (a) Calculate the exergy of the system. (b) Calculate the change in exergy for a process where the steam is heated at constant pressure until the volume doubles. (c) Calculate the change in exergy for a process where the steam isothermally expands until its volume doubles