Heat of Fusion and Vaporization Lecture Notes

Heat of Fusion and Vaporization

Overview

• Heat of fusion and heat of vaporization are measures of the energy required for a substance to undergo a phase change.

Heat of Fusion

• Definition: The amount of energy needed for a substance to melt or freeze under constant pressure.

• Specifically concerns the solid-liquid relationship.

• Solid to liquid is melting; liquid to solid is freezing.

Heat of Fusion of Water

• Water has a heat of fusion (Lf) of 334 \frac{J}{g}. This is a constant, intensive property.

• Meaning:

  • If you have 1 gram of liquid water, 334 Joules of energy must be released to freeze it.

  • If you have 1 gram of ice, 334 Joules of energy must be added to melt it.

Heat of Vaporization

• Describes the amount of energy needed for a substance to boil or condense under constant pressure.

• Water has a heat of vaporization of 2257 \frac{J}{g}.

• The heat of vaporization for water is significantly greater than the heat of fusion because the movement of particles is much higher in the gas phase compared to the liquid phase.

• To condense a gas, a lot of energy needs to be removed, more so than with a liquid or solid.

Comparison with Specific Heat Capacity

• Liquid water has a specific heat capacity of 4.18 \frac{J}{g \cdot ^{\circ}C}.

• Changing the temperature of liquid water is difficult due to its high specific heat capacity.

• The liquid to gas phase change involves a lot of particle motion, and high amounts of energy are needed to either separate them (boiling) or calm them down (condensing).

• Heat of vaporization is always higher than heat of fusion.

Time-Temperature Graph

• Illustrates the relationship between energy absorbed and resulting temperature for water (ice, water, water vapor).

• During a phase change, temperature does not change, as energy is used to weaken or strengthen intermolecular forces.

• Temperature is the average kinetic energy of particles (how fast they're moving).

• Phase change involves the distance between particles (closer or further apart).

• The energy isn't used to make particles move faster during phase change, but to change the distance between them.

• Melting Point: The temperature at which a substance melts or freezes.

• Boiling Point: The temperature at which a substance boils.

• During a phase change, the temperature remains constant because the energy being added is used to separate or strengthen intermolecular forces.

Latent Heat

• The energy that is not utilized to change the temperature. *Heat of fusion (Lf) and heat of vaporization (Lv) are examples of latent heat.

  • If you're doing some studying outside this class, you may see the heat of fusion be called latent heat of fusion. They mean the same thing.
    *Ice Cube Example:

  • An ice cube at 32 degrees Fahrenheit. Adding heat to it.

  • When it's becoming water, when it finishes becoming water, the temperature doesn't suddenly increase, it stays the same.
    *Ice Example:

  • You have the sample of ice at -5 degrees.

  • Increase the temperature by 5 degrees.

  • The resulting temperature is zero degrees.

  • The state of matter is still a solid, because you only absorbed enough energy to get to the melting point.

  • You haven't expended any energy to actually melt the substance, you have just got it to the temperature in which it can start to melt.

  • Once the phase change happens, the temperature are you at? Still Zero.

  • But what state of matter are you? Liquid.
    *Example:

  • If you have five grams of water at room temperature, 23 degrees, you cool it down to zero

  • It is water, liquid water at zero degrees

  • From there, excess energy needs to be removed from that water to actually transition it from a liquid to a solid.

  • So, it's the heat of fusion that needs to get added to the water at that state in order for it to change phase.

Molar Heat of Fusion and Vaporization

• Similar to heat of fusion/vaporization, but considers one mole of the substance.

• Molar mass: the amount of material needed to have one mole.

• Molar mass is an intensive property.

• Molar heat of fusion: energy needed to absorb or release for a solid-to-liquid phase change of one mole of a substance.

• Molar heat of vaporization: energy needed to be absorbed or released for a liquid-to-gas phase change of one mole of a substance.

• Water Example:

  • Heat of fusion of water is 334 \frac{J}{g}.

  • Multiply 334 \frac{J}{g} by water's molar mass (18.02) then divide by 1000 (to convert to kilojoules), resulting in 6.02 \frac{kJ}{mol}.

• Molar heat of fusion is more applicable with thermochemical stoichiometry.

• Molar heat of vaporization of water = 40.67 \frac{kJ}{mol}. You can find the value on the reference sheet

Converting Between Molar Heat and Specific Heat

• During a phase change, temperature does not change, so specific heat capacity (which includes temperature change) cannot be used.

Formulas

• Latent heat of fusion = \frac{\Delta H}{Molar \, mass}, where \Delta H is the change in enthalpy.

• The only difference between latent heat of fusion and latent heat of vaporation is the phase change involved.

Molar Mass

*Molar Mass = \frac{grams}{mole}

Latent heat of fusion = \frac{joules}{gram}

Kilojoules per mole = \frac{Molar \, Mass \cdot Latent \, heat \, of \, fusion}{1000}

*Ethanol Example
* Molar heat of fusion for ethanol is approximately 5.02 \frac{kJ}{mol}.
*To find the heat of fusion in \frac{J}{g} all you have to do is multiply it by 1000. So 5.02 \frac{kJ}{mol} becomes 5020 \frac{J}{mol}

Phase Change Diagram

• Pressure can influence the state of matter.

• Propane tanks Store liquid propane under pressure. When the valve is opened, you are relieving the sample of its pressure.

• By lowering the pressure, the particles can then spread out more and become a gas.

• Pressure and temperature dictate a state of matter.

Critical Point

• A temperature and pressure needed for a substance's liquid and gaseous phases to be indistinguishable from each other.

• At the critical point, it's hard to tell whether the volume is definite or indefinite.

• If a substance goes past the critical point it turns into a supercritical fluid.

• Fluid: ability to move (liquid or gas).

• Increasing temperature and pressure significantly results in a supercritical fluid.

• Supercritical fluid: A blend between liquids and gases.

Triple Point

• All three states of matter (solid, liquid, gas) coexist together in equilibrium.

• At a certain temperature and pressure, you will have all three states of matter being produced at the same time.

• Water at 1 ATM and 5 degrees Celsius is a liquid.

• To exist as solid, liquid, and gas all at the same time:

  • Low temperature, but just high enough to where it's not 100% a solid.

  • The pressure needs to be incredibly low

Triple Point of Water

• Temperature of 0.01 degrees Celsius (just above melting).

• Pressure of 0.006 atmospheres (incredibly low pressure).

• Results in all three states of matter existing in equilibrium.

• You may see water freezing, melting, boiling, and condensing all at the same time.

• Vacuum chambers allows a scientist to get rid of pressure.

Practice Problems

Key Considerations

• Phase changes

• Specific heat capacities at different states of matter.

Problem 1

Consider a 5.1 gram sample of solid gallium at room temperature. If the melting point of gallium is 30 degrees Celsius, how much energy in joules is needed to melt the sample of gallium? We're given the molar heat of fusion and the specific heat.

• Determining givens. So 5.1 gram sample. We're working with gallium, starting at room temperature. We got the melting point, and we're trying to solve for energy.

• Haven't even made it to the melting point yet. So we can determine how much energy is needed to get to the melting point. Since we're talking about a temperature change, we're going to use specific heat capacity.
  Find the energy needed to reach the melting point
  Q = mc\Delta T
  Q = (5.1 \, g) \cdot (0.37 \frac{J}{g \cdot ^{\circ}C}) \cdot (30^{\circ}C - 23^{\circ}C)
  Q = (5.1 \, g) \cdot (0.37 \frac{J}{g \cdot ^{\circ}C}) \cdot (7^{\circ}C)
  Q = 13.2 \, J

• This number is describing The amount of energy needed to be absorbed by this sample of solid gallium to get to the melting point.

• Once you get to the sample's melting point, what do you need to do? It might go through a phase change
  *Since there is no temperature change, we're going to use the latent heat of fusion.
  heat of fusion: 80. \frac{J}{°C}

• How to get the heat of fusion value? You need to use this format, \frac{kJ}{mol}
  to\frac{J}{°C}
  By using the following formula \frac{amount \, of \, energy \, * \, 1000}{Gallium's \, molar \, mass}
  Q = mL_f
  Q = (5.1 \, g) \cdot (80.3 \frac{J}{g})
  Q = 409.5 \, J

• You take both calculations and add them together to get the total calculations.

Therefore the total energy is 422.7 \, J

Problem 2

Consider an 11.7 gram sample of liquid water at room temperature. How much energy in kilojoules is needed to heat the sample of water from 23 to one zero five?
*You have to do multiple steps.

• The first step You want to start by writing, like, your first initial change in temperature
  *Not at the boiling point or melting point write now. That being said, we could just use specific heat capacity.
  Find the energy needed to reach the boiling point
  Q = mc\Delta T
  Q = (11.7 \, g) \cdot (4.18 \frac{J}{g \cdot ^{\circ}Cs}) \cdot (100^{\circ}C - 23^{\circ}C)
  Q = (11.7 \, g) \cdot (4.18 \frac{J}{g \cdot ^{\circ}C}) \cdot (77^{\circ}C)
  Q = 3766.67 \, J
  *So when we when this much energy is absorbed into this much water at room temperature, what is the new temperature right now?A hundred degrees.
  *What state of matter are we right now? We're still out of liquid. Yeah. Again, the process of boiling hasn't begun yet.
  Now it's time to find the energy needed to boil.
  Q = mL_v
  Q = (11.7 \, g) \cdot (2257 \frac{J}{g})
  Q = 26406.9 \, J
  *The next step is to see How much energy we need To reach that 105 goal, but it is very minor compared to others.
  Q = mc\Delta T
  Q = (11.7 \, g) \cdot (2.01 \frac{J}{g \cdot ^{\circ}}Cs) \cdot (105^{\circ}C - 100^{\circ}C)
  Q = (11.7 \, g) \cdot (2.01 \frac{J}{g \cdot ^{\circ}C}) \cdot (5^{\circ}C)
  Q = 117.6 \, J
  *Total energy = 3766.67 J + 26406.9 J + 117.6 J
  *Now you want to, combine them all and use the following format
  Final energy: \frac{30291}{1000}

Final Answer: \, 30.291 \, kJ