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Heat of vaporization is the quantity of heat used per unit mass to change the state of a substance. It is the heat required while a substance in liquid form changes its state to vapour or gas.
- The characteristics of a substance affect the heat vaporization capacity of that substance.
- A substance is present in three probable states, namely, gas, liquid, or solid. A substance may change its state from one to another after heating or cooling it.
- When a substance changes its state from liquid to gas or vapour, it is called vaporisation. Although, every substance has a specific melting or boiling point.
- The kinetic energy of steam is greater than that of fluid if the solutions of vapour and liquid states are compared.
- Heat of vaporization is the total amount of heat necessary to turn liquid into vapour without increasing the fluid's temperature.
Read Also: Difference between vaporization and evaporation
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Key Terms: state, substance, vaporization, energy, temperature, kinetic energy, entropy, enthalpy, boiling point.
Concept behind Heat of Vaporization
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While transforming water vapour into a liquid state, it is essential to take out the energy from gas. Even more, the amount of energy per unit mass required to transform water vapour is equivalent to the heat of vaporization.
- Apart from this, water’s heat of vaporization is the greatest known value. Thus we can call the heat of vaporization the amount of heat required to transform the liquid into gas.
- In this process, it is important to maintain the temperature of a liquid.
- The heat of vaporization is a kind of latent heat and calorie per gram is the unit of the same.
- Here, latent refers to remaining concealed or secret.
- Thus, latent heat is the extra heat required while transforming a substance’s state from solid to liquid.
Read Also: Heat Formula
Heat of Vaporization
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The amount of heat required for transforming a substance’s state from liquid to gas at its boiling points is called Heat of vaporisation. Notably, latent heat is strongly associated with a change in the state without any change in the temperature.
- The evaporation process of the water contains a cooling effect whereas condensation offers a heating effect.
- Thus, the heat of vaporization is the heat that is required to be absorbed to form vapours of a specific amount of liquid at a stable temperature.
- When we compare vapour solutions and liquid states, the kinetic energy of steam is greater than that of liquid.
- In the process of vaporization, a rise in energy enables liquid particles to evaporate by conquering intermolecular attractions.
- It is essential to have a specific quantity of heat to make physical changes happen and switch the states of a substance.
- This energy or heat used per unit mass at the time of vaporization is called the heat of vaporization or enthalpy of vaporization.
Formula for Heat of Vaporization
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The formula of heat of vaporization depending on enthalpy and entropy of the vaporization process is as follows:
\(H_v= {Q\over m}\)
Where,
- Hv = heat of vaporization
- Q = amount of heat, and
- m = mass of a given substance.
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Things to Remember
- The quantity of energy required while changing a substance from liquid to gas is the heat of vaporization.
- The heat of vaporization varies from substance to substance but is fixed for each substance.
- The formula for heat of vaporization is Hv=qm, where, Hv=heat of vaporization, q=heat, and m=mass of a given substance.
- Vaporization is the change of state, i.e., the state of a substance changes from liquid to vapour or gas. Even more, during the process of vaporization, the temperature remains unchanged or constant.
- In thermal equilibrium, both the vapour and liquid states of a substance co-exist. The boiling point of a substance is the temperature at which the vapour and liquid states of a substance co-exist.
Read Also: Three States of Matter
Sample Questions
Ques. Consider the heat of vaporization for H2O as 2357 J/g, what will be the total amount of energy required in the vaporization of 145 gm H2O? (2 Marks)
Ans. The formula for the heat of vaporization is Hv = q/m,
We can re-write the formula as, q = Hv × m
Given that, Hv=2357 J per gram, and m, i.e., the mass of a given substance is 145.
By substituting the given values in the formula, we get
q=2357×145
Thus, q=341765 J, which is the amount of heat required in the vaporization of 145 gm of H2O.
Ques. Calculate the energy required to vaporize 135 gm of water, when the heat of vaporization for the substance is 2257 J/g. (2 Marks)
Ans. To solve the given problem, we can re-arrange the formula for the heat of vaporization as:
q = Hv × m
q=2257Jg×135 g
Thus, q=305000 J is the required energy to vaporize 135 grams of water.
Ques. Calculate the mass of water vaporized at temperature 100 °C, and 15400 J of heat is applied. (2 Marks)
Ans. The problem contains the heat of vaporization for water, and to calculate mass, we can re-write the formula for heat of vaporization as
m = q/Hv
m=15400 J / 2257Jg-1
m=6.82 g
Ques. Calculate the heat (in joules) required to transform 25 gm of water into steam. Also, mention the heat required (in calories) for the same. (5 Marks)
Ans. Heat of vaporization of water = 2257 J/g = 540 cal/g
Part 1: To calculate the amount of heat required in joules:
Using the formula
q = Hv × m
Where, q = heat, m = mass, and Hv = heat of vaporization, we get
q = 2257 J/g × 25 g
q=56425 J
Part 2: To calculate the amount of heat required in calories:
q = Hv × m
q= 540 Cal/g × 25 g
q=13500 cal
Ques. Calculate the energy required to transform 50 gm of h2O into steam. Solve in terms of both, joules and calories. (4 Marks)
Ans. Given that, mass (m) = 50 gm,
We know that, the heat of vaporization of water = 2257 J/g
Substituting the given values in the formula, we get
q = Hv × m
q= 2257 × 50
q=112850 J=112.85 kJ
In calories: q=50×540
q=27000 cal=27 kCal
Thus it requires 112850 joules or 27 kcal of heat to transform 50 gm of H2O into vapor or gas.
Ques. What is the heat of vaporization of sulfur, if sulfur (in liquid state) vaporizes at 445°C and it takes 28125 joules to transform 20gm of sulfur (from liquid to gaseous state)? (3 Marks)
Ans. Given that m = 20 g and q=28125 J
Putting the values of q and m in the formula of heat of vaporization, we get
28125 J= Hv × 20
By solving the equation, with respect to Hv, we get
Hv = 28125 J / 20 g
Thus, Hv =1406.25 J/g
Therefore, the heat required to vaporize the given amount of sulfur is 1406.25 J/g.
Ques. Calculate the latent heat of substance with 5 kg mass, if the amount of heat required for a phase change is 300 kcal. (2 Marks)
Ans. Given that, q = 300 kcal, and m = 5 kg or 5000 g
The formula for latent heat is
L=q / m
L=300 / 5
L=60 kcal / kg.
Thus, the latent heat required is 60 kcal/kg.
Ques. At 20°C, a metal piece has a density of 60g. When immersed in a steam current at 100°C, 0.5g of steam condenses on the metal piece. If the latent heat of steam is 540 cal/g, what will be the specific heat of the piece of metal? (5 Marks)
Ans. Let, c = specific heat of the metal piece
Heat gained by the metal piece q=mcât
q=60×c×(100-20)
q=60×c×80 cal ……. (1)
The heat released by the steam,
q=mL
q=0.5×540 cal ……. (2)
By using the principle of mixtures, the heat given is the same as the heat taken.
By equating equation (1) and (2), we have
0.5×540=60×c×80
c=0.056 Cal/g °C
Thus, the specific heat value required is 0.056 Cal/g °C.
Ques. What is essential to change the solid state of water to liquid or gas (vapour)? (2 Marks)
Ans. The state of vaporization, pressure, and temperature are required to transform any substance from a liquid state to a vapour or gaseous state. The heat of vaporization helps a substance transform its state from liquid to gaseous.
When a substance in a solid state reaches its melting point, the state changes to a liquid. Also, if a solid-state substance reaches boiling point, it gets transformed into a gaseous state. Thus, whenever a solid substance reaches its melting point it will result in a liquid state and this liquid further heated up to a boiling point will convert liquid water into a gaseous state or vapour.
Ques. A liquid lead chunk of weight 10g at the temperature of 1750°C requires 8580 J of heat to transform into the gaseous lead. What will be the latent heat of vaporization of the given substance? (2 Marks)
Ans. A liquid state is being transformed into a gaseous state, so there is a change in phase. Substitute the given values in the formula of the heat of vaporization, we have
q=mL
rm8580 J = (0.010 kg)L
L = 858 kJ/kg
Thus, the heat required to vaporize a given amount of lead is 858 joules per kg.
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