Physics Lesson 13.3.5 - Latent Heat. Specific Latent Heat of Fusion and Vaporization

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Welcome to our Physics lesson on Latent Heat. Specific Latent Heat of Fusion and Vaporization, this is the fifth lesson of our suite of physics lessons covering the topic of Absorption of Heat, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Latent Heat. Specific Latent Heat of Fusion and Vaporization

Since there is no change in temperature during the phase change, we cannot use anymore the formula Q = m × c × ΔT to calculate the heat supplied to or removed from a substance during such a process, as ΔT = 0 and therefore, the value of Q would result zero as well. It means we must find other ways to calculate the heat during this process.

It is clear that the substance gains or loses heat during a phase change regardless its temperature remains constant. For this reason, this kind of heat - often forgotten to be considered during calculations - is known as latent (hidden) heat.

By definition, latent heat of fusion, Q_f, is the amount of heat absorbed by a substance at the melting temperature in order to melt it completely.

Similarly, latent heat of vaporization, Q_v, is the amount of heat absorbed by a substance at the boiling temperature in order to evaporate it completely.

Not all substances require the same amount of heat energy to change their state. Therefore, similarly as in the case of specific heat capacity, we introduce a new quantity known as the specific latent heat, L, which represents the amount of heat per kilogram mass needed to change the phase of a material. By definition,

Specific latent heat of fusion L_f is the amount of heat supplied to 1 kg of a substance in its melting temperature in order to make it melt completely.

Similarly, specific latent heat of vaporization L_v is the amount of heat supplied to 1 kg of a substance in its boiling temperature in order to make it evaporate completely.

Both specific latent heats are measured in Joules per kilogram, [J/kg]. The equation of latent heat for both cases therefore is

Q_f = m × L_f

and

Q_v = m × L_v

Example 3

What is the amount of heat absorbed by 200 g of solid lead at 600 K? Melting temperature of lead is 327°C and its specific latent heat of fusion is 22 900 J/kg.

Solution 3

600 K correspond to 327°C because T(K) = t°C + 273°. This is equal to the melting temperature of lead. Therefore, we have to use only the equation

Q_f = m × L_f

to calculate the heat absorbed by lead during this stage. Also, we have to write the mass of lead in kilograms, i.e. m = 0.3 kg. Thus, we obtain after substituting the known values

Q_f = 0.3 kg × 22900 J/kg
=6370 J

for the heat absorbed by lead during the melting stage.

We can combine the formulae Q = m × c × Δ t and Q = m × L to calculate the amount of heat absorbed when the material is not at the melting or boiling temperature. Let's consider an example to clarify this point.

Example 4

What is the amount of heat absorbed by 50 g ice at -20°C in order to melt it completely? Take the melting (freezing) temperature of ice (water) equal to 0°C. Specific heat capacity of ice is 2100 J/kgK and the latent heat of fusion for ice is 334 000 J/kg.

Solution 4

The material passes into two stages during this event:

The ice is "heated" from -20°C to 0°C. During this process, we must use the formula Q = m × c × Δ T.
The ice is melted when it reaches the temperature 0°C. During this process, we must use the formula Q = m × L_f.

Given that m = 50 g = 0.05 kg, c = 4200 J/kgK, t₁ = -20°C, t₂ = 0°C and L_f = 334 000 J/kg, we write:

Q_total = Q₁ + Q₂
= m × c × (t₂ - t₁ ) + m × L_f
= 0.05 × 2100 × [0 - (-20)] + 0.05 × 334000
= 2100 J + 16700 J
= 18800 J

The graph below represents the relationship between heat and temperature in all possible stages.

Physics Tutorials: This image provides visual information for the physics tutorial Absorption of Heat

In part I of the graph, the substance is in solid state. We must use the formula Q = m × cs × ΔTs (s stands for solid) to calculate the heat absorbed by it during this stage.

In part II of the graph, the substance is melting. We must use the formula Q = m × L_f to calculate the heat absorbed by it during this stage.

In part III of the graph, the substance is in liquid state. We must use the formula Q = m × cl × ΔTl (l stands for liquid) to calculate the heat absorbed by it during this stage.

In part IV of the graph, the substance is boiling (evaporating at the highest rate). We must use the formula Q = m × L_v to calculate the heat absorbed by it during this stage.

In part V of the graph, the substance is in gaseous state. We must use the formula Q = m × cg × ΔTg (g stands for gas) to calculate the heat absorbed by it during this stage.

Remark!

The specific heat capacities are not equal for the same material. For example, for water the value of specific heat capacity is 4186 J/kgK while for ice or water vapour, it is 2100 J/kgK. This is true for latent heats as well. For example, latent heat of fusion for ice is 334 000 J/kg and the latent heat of vaporization for water is 2 224 000 J/kg.
If mass is given in grams and heat in calories, we can use the conversion factor 1 cal/g × °C = 4180 J/(kg × °C) in order to deal with smaller numbers.
We can find the heat released by a hot object when it cools down by using the same procedure as when it is heated up. The only difference is that the heat value will result a negative number because t₂ < t₁ and therefore, Δt = t₂ < t₁ < 0.

Example 5

A 200 ml glass full of water initially at the environment temperature (20°C), is placed inside a refrigerator whose temperature is -10°C. Given that water has a density of 1000 kg/m³ and 1 ml = 1 cm³, calculate the amount of heat (in calories) released by water during this process. Take c_water = 1 cal/gK, cice = 0.5 cal/g and L_fusion = 80 cal/g.

Solution 5

Let's convert the clues into the required units first. We have

V = 200 ml = 200 cm³
t₁ (water) = 20°C
t₂ (ice) = -10°C
ρ_water = 1000 kg/m³ = 1 g/cm³
c_water = 1 cal/gK
c_ice = 0.5 cal/g
L_fusion = 80 cal/g
Q_{total released} = ?

First, we must work out the mass (in grams) of water, as it is needed during the solution. We have

m = ρ × V
= 1 g/cm ³ × 200 cm³
= 200 g

In this problem, water experiences three stages:

1) It cools down from 20°C to 0°C (the minimum temperature in which water can exist in liquid state). During this process, we must use the equation

Q₁ =m × c_w × ∆t_w
= m × c_w × (t_melting - t_1(water) )
= 200 g × 1 cal/g∙°C × (0 - 20)°C
= -4000 cal

2) Then, water freezes without any change in temperature. During this process, we must use the equation

Q₂ = m × L_f
= -200 g × (-80)cal/g
= -1600 cal

3) Finally, the ice cols down from 0°C to -10°C. During this process, we must use the equation

Q₃ = m × c_w × ∆t_w
= m × c_ice × (t_{2 (ice)} -t_melting )
= 200 g × 1 cal/g × °C × (-10 - 0)°C
= -2000 cal

Therefore, we obtain for the total energy

Q_total = Q₁ + Q₂ + Q₃
= -4000 cal - 1600 cal - 2000 cal
= -7600 cal

This means the water loses 7600 cal of heat energy during this process.

You have reached the end of Physics lesson 13.3.5 Latent Heat. Specific Latent Heat of Fusion and Vaporization. There are 5 lessons in this physics tutorial covering Absorption of Heat, you can access all the lessons from this tutorial below.

More Absorption of Heat Lessons and Learning Resources

Thermodynamics Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
13.3	Absorption of Heat
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
13.3.1	Absorption of Heat. Good and Bad Absorbers of Heat
13.3.2	Joule or Calorie?
13.3.3	States of Matter
13.3.4	Change of State
13.3.5	Latent Heat. Specific Latent Heat of Fusion and Vaporization

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