Question

A system contains H2O at P=150 kPa, v= 1.5 m^3 and a specific volume of 0.5 m^3/kg (state 1). It is heated at constant volume until the temperature reacues 160 celcius

What is the temperature at state 1 ?

Answer 1

For water specific heat capacity at constant volume=Cv=74.539J/K mol

mass of water=(1/specific volume)*volume=(1/0.5 kg/m^3)*1.5 m^3=3 kg

(m,)mole of water=mass of water/molar mass of water=(3*10^3g)/(18g/mol)=166.67 mol

T=T2-T1=change in temperature

T1=?

T2=160C=160+273=433K

Heat provided at constant volume =Q=(m)*Cv*T=166.67 mol*(74.539J/K mol)*(433K-T1)............(1)

Maximum density of water at 4 degree C(=273+4) or 277K=1000 kg/m3

Or, it has maximum mass per unit volume at 277K

Or, water has lowest volume at 277K

So maximum volume expansion of water possible by providing heat Q=∆V=V2-V1

V2=0.001 m3/kg

V1=0.5m3/kg

∆V=0.499 m3/kg

P=150KPa=150 *1000 Pa *1/101325 Pa/atm=1.48 atm

Maximum Pressure –volume produced=p∆V=1.48 atm *0.499 m3/kg =0.739 m3 atm/kg=0.739 m3 atm/kg *1000L/m3=739 atm L/kg

For the change of temperatute from T1 to 160C, maximum pressure volume work possible=3kg*739 atm L/kg *(8.314 J/0.082 atmL)=18432.14 J=18.432 kj

AS heat is supplied at constant volume so all heat is converted into internal energy or used up in change of temperature.

Q= 166.67 mol*(74.539J/K mol)*(433K-T1)=18432.14J

433K-T1=1.484

T1=431.52 K=431.51-273=158.52C answer