Why is enthalpy of water vapour and liquid different
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the enthalpy of water vapour and liquid different by following ways
for liquid
Specific enthalpy of saturated water - hf - can be obtained from tables as above. The value depends on the pressure.
For saturated water at standard atmosphere - 2) -the specific enthalpy - hf - is 419 kJ/kg.
At standard atmosphere - 1 bar (14.7 psi) - water starts boiling at 100 oC (212 oF).
The specific enthalpy of water (in SI units) can be calculated from:
hf = cw (tf - t0) (3)
where
hf = enthalpy of water (kJ/kg)
cw = specific heat water (4.19 kJ/kg.oC)
tf = saturation temperature (oC)
t0 = refer temperature = 0 (oC)
for evaporate
The energy to evaporate a certain amount of water can be calculated as
Q = he m (4b)
where
Q = evaporation energy (kJ)
m = mass of water (kg)
The energy to evaporate 5 kg of water at atmospheric pressure can be calculated as
Q = (2257 kJ/kg) (5 kg)
= 11285 kJ
hope it helps you
for liquid
Specific enthalpy of saturated water - hf - can be obtained from tables as above. The value depends on the pressure.
For saturated water at standard atmosphere - 2) -the specific enthalpy - hf - is 419 kJ/kg.
At standard atmosphere - 1 bar (14.7 psi) - water starts boiling at 100 oC (212 oF).
The specific enthalpy of water (in SI units) can be calculated from:
hf = cw (tf - t0) (3)
where
hf = enthalpy of water (kJ/kg)
cw = specific heat water (4.19 kJ/kg.oC)
tf = saturation temperature (oC)
t0 = refer temperature = 0 (oC)
for evaporate
The energy to evaporate a certain amount of water can be calculated as
Q = he m (4b)
where
Q = evaporation energy (kJ)
m = mass of water (kg)
The energy to evaporate 5 kg of water at atmospheric pressure can be calculated as
Q = (2257 kJ/kg) (5 kg)
= 11285 kJ
hope it helps you
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