Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.25 kg of water decreased from 125 °C to 24.5 °C.

Melting point: 0 degrees C

Boiling Point: 100 degrees C

Delta H(fus): 6.01 kJ/mol

Delta H(vap): 40.67 kJ/mol

cp(s): 37.1 J/mol x degrees C

cp(l): 75.3 J/mol x degrees C

cp(g): 33.6 J/mol x degrees C

Answer 1

The process takes place in a number of steps :

Step 1) E1=energy change when temperature of water vapours decreased from 125 deg C to 100 deg C(b.pt)

E1=mass of water* Cp(g)*T=(1.25*1000g)*(4.184J/g .deg C)*(100-125)degC=-130750J(negative sign for heat released)

Step 2: E2=energy change for phase transition from gaseous to liquid

E2=Hvap*mol of water=Hvap*(mass of water/molar mass of water)=-40.67 KJ/mol*(1.25*1000g)/(18.015g/mol)=-2821.953 KJ

Step 3)E3 =energy change when temperature of water decreased from 100 to 24.5 deg C

=(1.25*1000g)*(4.184J/g .deg C)*(24.5-100)degC=-394865 J

total energy change (released)=(-130750J)+(-2821.953 KJ)+(-394865 J)=-3347568J=-3347.568KJ

total energy change=-3347.568KJ