# What is Enthalpy of Vaporization? ##### Joseph Hebert
Owner at - HVAC Buster

Joseph is an HVAC technician and a hobbyist blogger. He’s been working as an HVAC technician for almost 13 years, and he started blogging just...Read more

In thermodynamics, the enthalpy of vaporization (also known as the heat of vaporization or latent heat of vaporization) is the amount of energy required to turn a given quantity of liquid into a gas at a given constant temperature. The enthalpy of vaporization is usually expressed in terms of the mole and is often reported in kJ/mol or J/mol.

The enthalpy of vaporization, also known as the heat of vaporization or evaporative heat, is the amount of energy that must be added to a liquid in order to transform it into a gas. The enthalpy of vaporization is usually expressed in terms of the mole fraction of the substance that is being vaporized. For example, water has an enthalpy of vaporization of 2260 kJ/mol, which means that it takes 2260 kJ of energy to transform one mole (6.023 x 10^23 molecules) of water into steam.

The enthalpy of vaporization is an important property because it can be used to calculate the amount of energy required to Vaporize a given mass or volume of a liquid. This information is useful for many industrial and engineering applications such as: boiler design, refrigeration, and power generation.

## Enthalpy of Vaporization of Water

The enthalpy of vaporization, also known as the heat of vaporization or latent heat of vaporization, is the amount of energy required to turn a liquid into a gas. The enthalpy of vaporization is usually expressed as a temperature, such as 100°C. For water, this value is 2260 kJ/kg.

This value can be used to calculate the amount of energy required to turn 1 kg of water into steam at 100°C. To do this, we simply multiply the enthalpy of vaporization by the mass of water: 2260 kJ/kg x 1 kg = 2260 kJ

So, it takes 2260 kJ of energy to turn 1 kg of water into steam at 100°C. This value can be used to calculate the amount of energy required to turn any mass or volume of water into steam at that temperature.

## How to Calculate Enthalpy of Vaporization

The enthalpy of vaporization (∆Hvap) is the heat required to convert 1 mol of a liquid into its gaseous form at a given temperature. The SI unit for ∆Hvap is kJ/mol. The constant-temperature molar heat of vaporization (Δ vap H°), also called latent heat of vaporization or heat of evaporation, is usually expressed in J/mol or kJ/mol.

It is the enthalpy change that accompanies the phase transition from liquid to gas at a given constant temperature. At the boiling point, both the liquid and gas phases are in equilibrium with each other and ∆Hvap = Δ vap H°. To calculate ∆Hvap, we first need to determine Δ vap H°.

This can be done using either experimental data or theoretical calculations. If experimental data is available, Δ vap H° can be calculated using the following equation: Δ vap H° = (moles of gas x molar mass of gas x R x T)/(moles of liquid x density of liquid)

Where R is the universal gas constant (8.31447 J/(K·mol)), T is the boiling point temperature in Kelvin, and density ofliquid is the density of the liquid at its boiling point. If experimental data is not available, we can use theory to calculate an estimate for Δ vap H° . For instance,the ideal gas law tells us that PV = nRT , where P is pressure, V is volume, nis numberof particles (in this case, molecules), and Tis temperature .

We can rearrange this equationto solve for V: V= nRT/P . If we substitute P with PA , where Ais area ,we get an expressionfor volume that only depends on temperature and numberof particles: V= nRT/PA . Now let’s assume that 1 molof our substance occupies 22.4 L when it’s agas at STP (standard conditions): 22.4 L= RT/PA .

Rearranging once again gives us anexpression for area: A= RT/(22.4L) . We cansubstitute A back into our original equationto get an expression that only depends onnumberof particles and temperature: PV =nRT/(22..4L) .

## Latent Heat of Vaporization of Water in J/Kg

Did you know that water has a latent heat of vaporization? This is the amount of heat required to change one gram of water from a liquid to a gas. The latent heat of vaporization for water is 2,260 joules per kilogram.

That means it takes a lot of energy to turn water into steam! This property of water is what makes it such a good cooling agent. When water evaporates, it absorbs a lot of heat from its surroundings.

This is why we perspire when we get too hot – our bodies are trying to cool themselves down by sweating! The latent heat of vaporization also plays an important role in the climate. Water vapor in the atmosphere helps trap heat and keep the Earth warm enough to support life.

Without this greenhouse effect, our planet would be much colder than it is today.

## Enthalpy of Vaporization of Water at 25 C

The enthalpy of vaporization of water at 25 C is 40.65 kJ/mol. This means that it takes 40.65 kJ/mol to convert 1 mol of liquid water at 25 C to 1 mol of steam at 25 C. The enthalpy of vaporization is sometimes also referred to as the heat of vaporization or the latent heat of vaporization. The enthalpy of vaporization is an important thermodynamic property because it is used to calculate the amount of energy required to convert a liquid into a gas.

In other words, it tells us how much energy must be added to a liquid in order to Vaporize it. The higher the enthalpy of vaporization, the more energy that must be added, and vice versa. Water has a very high enthalpy of vaporization compared to most other substances.

For example, the enthalpy of vaporization for ethanol (C2H5OH) is only 38 kJ/mol, while for methanol (CH3OH) it is only 33 kJ/mol. This means that it takes much less energy to Vaporize these alcohols than it does water. The reason why water has such a high enthalpy of vaporization is because its molecules are held together by strong hydrogen bonds.

When water molecules are heated, they break apart these bonds and turn into steam molecules, which requires quite a bit of energy.

## What is Enthalpy of Vaporization Give Examples?

The enthalpy of vaporization is the amount of heat that must be added to a liquid at its boiling point to produce 1 mole of vapor. The enthalpy of vaporization is usually expressed in kJ/mol or J/mol and is often called the heat of vaporization. For example, the enthalpy of vaporization of water at 100°C is 2260 kJ/mol.

This means that it takes 2260 kJ of energy to convert 1 mole of water from a liquid to a gas at 100°C.

## What is Enthalpy of Vaporization Class Ix?

In thermodynamics, the enthalpy of vaporization (also known as the heat of vaporization or latent heat of vaporization) is the amount of energy required to turn a given mass of liquid into a gas at a given temperature. The enthalpy of vaporization is usually expressed in terms of the specific heat (c), which is the amount of energy required to raise one gram of liquid by one degree Celsius. The enthalpy of vaporization is an important concept in many areas of physics and engineering, especially when dealing with steam power plants and refrigeration systems.

It can also be used to calculate the equilibrium constant for a chemical reaction.

## What is Enthalpy of Water Vaporization?

The enthalpy of water vaporization is the heat required to convert liquid water into gaseous water vapor. The value can be either positive or negative, depending on the direction of the process. If the process is exothermic (releases heat), then the enthalpy is negative; if it is endothermic (absorbs heat), then it is positive.

The standard enthalpy of vaporization (ΔH vap ) has a value of 2260 kJ/mol at 25°C. This means that it takes 2260 kJ/mol of energy to convert 1 mol of liquid water into 1 mol of gaseous water vapor at 25°C and 1 atmosphere pressure. The molar enthalpy of vaporization is sometimes also referred to as the heat of vaporization or evaporation.

The ΔH vap values for other temperatures can be calculated using the following equation: ΔH vap(T2) = ΔH vap(T1) + integral(C p dT) T1

## How Does Enthalpy of Vaporization Work?

When a liquid is heated, the molecules begin to move faster. Eventually, the molecules have enough energy to overcome the intermolecular forces that are holding them together in the liquid state. At this point, they vaporize, or turn into a gas.

The enthalpy of vaporization is the amount of heat that must be added to a liquid at its boiling point to vaporize it. The enthalpy of vaporization can be thought of as the “heat of evaporation.” It’s usually expressed in joules per mole (J/mol) or kilocalories per mole (kcal/mol).

The molar mass is the weight of one mole of a substance, and it’s used to convert between J/mol and kcal/mol. The enthalpy of vaporization is different for every substance. For example, water has a much higher enthalpy of vaporization than ethanol.

This means that it takes more heat to turn water into steam than it does to turn ethanol into vapors. The enthalpy of vaporization is important in many applications. For instance, power plants use huge amounts of water to create steam that turns turbines and generates electricity.

If the water didn’t have a high enough enthalpy of vaporization, it wouldn’t be able to turn into steam quickly enough and the power plant would shut down. Enthalpies of Vaporization for some common liquids: Substance Enthalpy of Vaporization (kJ/mol)

Water 40.65 Ethanol 38.55

## Conclusion

In short, the enthalpy of vaporization is the heat required to turn a liquid into a gas. The higher the boiling point of a substance, the more energy it takes to vaporize it. Joseph is an HVAC technician and a hobbyist blogger. He’s been working as an HVAC technician for almost 13 years, and he started blogging just a couple of years ago. Joseph loves to talk about HVAC devices, their uses, maintenance, installation, fixing, and different problems people face with their HVAC devices. He created Hvacbuster to share his knowledge and decade of experiences with people who don’t have any prior knowledge about these devices.