I was wondering if you could clarify this question.
How much thermal energy is needed to change 100 g of ice at -20 degrees celsius into steam at 110 degrees celsius?
Q = mLv (Lv = Specific latent heat of vapourization)
I can't find the Lv for ice. Is this a two step question where I need to calculate the Lf of water to ice and then add Q2(mLv)?
you need the heat req to heat the ice from -15C to 0C
then the heat to melt the ice at 0C
then to warm the water from 0C t0 100C
then to change the water to steam at 100C
then to heat the steam from 100C t0 110C.
Thermal properties for water:
Maximum density at 4 oC - 1000 kg/m3, 1.940 slugs/ft3
Specific Weight at 4 oC - 9.807 kN/m3, 62.43 Lbs./Cu.Ft, 8.33 Lbs./Gal., 0.1337 Cu.Ft./Gal.
Freezing temperature - 0 oC (Official Ice at 0 oC)
Boiling temperature - 100 oC
Latent heat of melting - 334 kJ/kg
Latent heat of evaporation - 2257 kJ/kg
Critical temperature - 380 oC - 386 oC
Critical pressure - 221.2 bar, 22.1 MPa (MN/m2)
Specific heat water - 4.187 kJ/kgK
Specific heat ice - 2.108 kJ/kgK
Specific heat water vapor - 1.996 kJ/kgK
Thermal expansion from 4 oC to 100 oC - 4.2x10-2 (Note! - volumetric temperature expansion of water is not linear with temperature)
Bulk modulus elasticity - 2.15 x 109 (Pa, N/m2)
Lf i
Yes, you're correct! To calculate the thermal energy needed to change a substance from one state to another, you need to consider both the specific latent heat of fusion (Lf) and the specific latent heat of vaporization (Lv). In this case, since you are changing ice at -20 degrees celsius to steam at 110 degrees celsius, you'll need to calculate the thermal energy required for both steps.
The specific latent heat of fusion (Lf) represents the heat energy needed to change a substance from a solid to a liquid state. The specific latent heat of vaporization (Lv) represents the heat energy needed to change a substance from a liquid to a gas state.
First, let's find the specific latent heat of fusion (Lf) for water. The value for Lf of water is 334 J/g.
To change 100 g of ice at -20 degrees celsius to water at 0 degrees celsius, you'll need to calculate the thermal energy required using the equation Q = mLf.
Q1 = 100 g x 334 J/g = 33400 J
Next, you need to find the specific latent heat of vaporization (Lv) for water. The value for Lv of water is 2260 J/g.
To change 100 g of water at 0 degrees celsius to steam at 100 degrees celsius, you'll need to calculate the thermal energy required using the equation Q = mLv.
Q2 = 100 g x 2260 J/g = 226000 J
Finally, to find the total thermal energy needed, add Q1 and Q2:
Total Q = Q1 + Q2 = 33400 J + 226000 J = 259400 J
Therefore, the total thermal energy needed to change 100 g of ice at -20 degrees celsius into steam at 110 degrees celsius is 259400 J.