Question

# Based on the thermodynamic properties provided for water, determine the energy change when the temperature of...

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

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

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