Question

Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 300°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 3560 kJ and kinetic and potential energy effects are negligible.

Determine the final volume, in m3, and the work for the process, in kJ.

Answer 1

The final volume is 3.43m^3 and the work for the process is 390kJ

Step-by-step explanation:

Quantity of heat (Q) = mc(T2-T1)

Q = 3560kJ = 3560×1000 = 3560000J

m (mass of water) = 5kg

c (specific heat capacity of water) = 4200J/kg°C

T1 (initial temperature of water) = 300°C

3560000 = 5×4200(T2-300)

T2-300 = 3560000/21000

T2-300 = 169.5

T2 = 169.5+300 = 469.5°C = 469.5+273K = 742.5K

From ideal gas equation

P1V1 = nRT1

V1 = nRT1/P1

number of moles of water = mass/MW = 5/18 = 0.278mole

R = 8314.34m^3.Pa/kgmol.K

T1 = 300°C = 300+273 = 573K

P1 = 5bar = 5×10^5N/m^2

V1 = 0.278×8314.34×573/5×10^5 = 2.65m^3

From the general gas equation,

P1V1/T1 = P2V2/T2

P1 = P2 = 5bar

Therefore, V1/T1 = V2/T2

V2 (final volume) = V1T2/T1 = 2.65m^3 × 742.5K/573K = 3.43m^3

Work = P(V2 - V1) = 5×10^5N/m^2(3.43m^3 - 2.65m^3) = 5×10^5N/m^2 × 0.78m^3 = 3.9×10^5Nm = 3.9×10^5J = 3.9×10^5/1000 = 390kJ