In this notebook, we explore how much energy is needed to vaporize (from microwaves) a certain quantity of water.
The specific heat of water is $C_P$ = 4185.5 J/(kg⋅K). Given a initial temperature of 15°C
In [1]:
T_init = 15 # °C
T_vapo = 100 # °C
C_P = 4185.5 # J/(kg.K)
m_water = 1 # kg
In [2]:
nrj_100 = m_water * C_P * (T_vapo - T_init)
print('Energy [J] required to heat the water up to 100°C : {} J'.format(nrj_100))
Enthalpy of Vaporization is the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure. For water at its normal boiling point of 100 ºC, the heat of vaporization is 2257 kJ/kg : this is the amount of energy required to convert 1 g of water into 1 g of vapor without a change in temperature.
In [3]:
C_v = 2257e3 # J/kg
nrj_vapor = m_water * C_v
print('Energy [J] required to convert {} kg of liquid water into vapor : {} J'.format(m_water, nrj_vapor))
In [4]:
print('Total energy required to vaporize {} kg of water at {}°C into vapor : {} J'.format(m_water, T_init, nrj_100+nrj_vapor))
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