A rigid, well-insulated container holds 0.5 m³ saturated water vapor at 200°C and 0.1 m³ solidsilver at 20°C, initially separated by a thermally insulating insert. Suddenly, the thermallyinsulating insert is removed and heat transfer occurs between the water and the silver.What is the final temperature of the water and silver after sufficiently long period of time.How much entropy is generated during this thermalization process?

It is given that thermalization process occurs in rigid and well insulated container. Given data are as follows: Volume of saturated water vapor Vw = 0.5 m3, Temperature of saturated water vapor Tw = 200oC, Volume of solid silver Vs = 0.1 m3, and Temperature of solid silver Ts = 20oC.At 200oC saturated condition form steam tableVg = 0.127222m3/Kg. Then mass of water will be given as, Mass of water mw=VwVg=0.5m30.127222m3Kg=3.93014Kg Taking the density of silver = 10497 Kg/m3.Mass of silver will be given as, Mass of silver Ms=Density×Vs=10497Kgm3×0.1m3=1049.7Kg Take specific heat capacity of silver Cp,s = 0.235 KJ/(Kg.oC) and specific heat capacity of saturated water vapor Cp,w = 1.996 KJ/(Kg.oC). Assume Tf be the final temperature of the system. It is given that thermally insulated insert is removed suddenly, so Change in internal energy of the system will be zero. Change in internal energy=0mwCp,w(Tf-200)+msCs,w(Tf-20)=03.93014Kg×1.996KJKg.oC×(Tf-200)+1049.7Kg×0.235KJKg.oC×(Tf-20)=07.84456Tf-1568.91189+246.6795Tf-4933.59=0254.52406Tf=6502.50189Tf=25.548oC It is known that entropy generated is given by the formula as shown below, Entropy ∆S=msCp,sLnTfTs+mwCp,wLnTfTw=1049.7Kg×0.235KJKg.oCLn25.548+273.1520+273.15+3.93014Kg×1.996KJKg.oCLn25.548+273.15200+273.15=4.6249KJoC-3.5541KJoC=1.0708KJoC Answer: Final temperature of the system is 25.548oC and entropy generated during this thermalization process is 1.708 KJ/(oC).