m1and m2can be found by using the formula: vVmState 1 is saturated vapor at 100kPa; the relevant properties can be found in table A-5. v1= vg at 100 kPa= 1.6941 m3/kgState 2 is at 300kPa and 200°C and is therefore a superheated vapor; the relevant properties can be found in table A-6.v2= 0.71643 m3/kg The container is a rigid tank and the volume V stays constant at 2 m3during this process.kgkgmmvVm791619558.2/0.7164323322kgkgkgmmmi81.61105155180568.182.7916195512(b) By using the first law relation for a transient process, the enthalpy of the supplied steam hsupcan be found.𝑄𝑐𝑣− 𝑊𝑐𝑣= ∑ 𝑚𝑒ℎ𝑒𝑒𝑥𝑖??− ∑ 𝑚𝑖ℎ𝑖𝑖𝑛𝑙𝑒??+ (𝑚2𝑢2− 𝑚1𝑢1)𝑐𝑣(1)Energy leaves the system during cooling, therefore, Qcv= –700 kJ.There is no work done on/by the system, Wcv= 0.Steam only enters the tank, so me= 0 and hi= hsup.kgkgmmvVm180568.1/6941.123311
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