How to Find Energy Required to Vaporize: A Comprehensive Guide

Vaporization is a fascinating process that occurs when a substance changes from its liquid state to a gaseous state. Whether it’s water boiling on a stovetop or a solvent evaporating from a surface, understanding the energy required for vaporization is crucial in various fields, including chemistry, physics, and engineering. In this article, we will explore the physics behind vaporization and learn how to calculate the energy required to vaporize a substance.

The Physics Behind Vaporization

The Role of Heat in Vaporization

When a substance undergoes vaporization, it absorbs energy in the form of heat. This heat energy overcomes the intermolecular forces holding the molecules together in the liquid state, allowing them to break free and transition to the gaseous state. The amount of heat required for this phase change is known as the heat of vaporization.

The Concept of Latent Heat

The heat of vaporization is also referred to as latent heat. “Latent” means hidden, and in the context of vaporization, it refers to the heat energy that is absorbed or released during a phase change without causing a change in temperature. This is because the absorbed heat is used to break the intermolecular bonds, rather than increasing the kinetic energy of the molecules.

Energy Transfer during Vaporization

During vaporization, energy is transferred from the surroundings to the substance being vaporized. This energy transfer depends on factors such as the substance’s specific heat, heat capacity, and the heat source’s temperature. Understanding these factors allows us to calculate the energy required for vaporization accurately.

How to Calculate the Energy Required to Vaporize

Understanding the Formula

To calculate the energy required to vaporize a substance, we can use the formula:

$Q = m cdot Delta H_v$

Where:
– $Q$ represents the energy required for vaporization (in joules).
– $m$ is the mass of the substance (in grams).
– $Delta H_v$ is the heat of vaporization (in joules per gram).

Required Parameters for Calculation

To use the formula, we need to determine the mass of the substance being vaporized and the heat of vaporization specific to that substance. The heat of vaporization can be found in reference books or online databases for different substances.

Step-by-Step Calculation Process

Let’s go through a step-by-step calculation process using the formula mentioned above:

Step 1: Determine the mass of the substance being vaporized (m) in grams.

Step 2: Find the heat of vaporization $( Delta H_v$ ) of the substance in joules per gram.

Step 3: Multiply the mass of the substance by the heat of vaporization to find the energy required for vaporization (Q) in joules.

Practical Examples of Calculating Energy Required to Vaporize

Example 1: Calculating Energy Required to Vaporize Water

Let’s say we want to calculate the energy required to vaporize 100 grams of water. The heat of vaporization for water is approximately 40.7 kJ/mol. To convert this value to joules per gram, we need to divide it by the molar mass of water, which is about 18.015 g/mol.

Given:
– Mass of water (m): 100 grams
– Heat of vaporization of water $( Delta H_v$ ): 40.7 kJ/mol / 18.015 g/mol

$Q = m cdot Delta H_v = 100 , text{g} times left(frac{40.7 , text{kJ/mol}}{18.015 , text{g/mol}}right)$

Calculating the value:

$Q approx 226.2 , text{kJ}$

Hence, approximately 226.2 kJ of energy is required to vaporize 100 grams of water.

Example 2: Calculating Energy Required to Raise Temperature

Image by Steffen 962 – Wikimedia Commons, Wikimedia Commons, Licensed under CC0.

Let’s consider a different scenario where we want to calculate the energy required to raise the temperature of 50 grams of water from 25°C to its boiling point at 100°C. In this case, we need to account for both the energy required to increase the temperature (using the specific heat) and the energy required for vaporization.

Given:
– Mass of water (m): 50 grams
– Specific heat of water: 4.18 J/g°C
– Initial temperature (T1): 25°C
– Final temperature (T2): 100°C
– Heat of vaporization of water $( Delta H_v$ ): 40.7 kJ/mol / 18.015 g/mol

Step 1: Calculate the energy required to raise the temperature:
$Q_{text{temperature change}} = m cdot C cdot Delta T$
$= 50 , text{g} times 4.18 , text{J/g°C} times (100 - 25) , text{°C}$

Calculating the value:
$Q_{text{temperature change}} = 17450 , text{J}$

Step 2: Calculate the energy required for vaporization:
$Q_{text{vaporization}} = m cdot Delta H_v$
$= 50 , text{g} times left(frac{40.7 , text{kJ/mol}}{18.015 , text{g/mol}}right)$

Calculating the value:
$Q_{text{vaporization}} approx 113.6 , text{kJ}$

Step 3: Find the total energy required:
$Q_{text{total}} = Q_{text{temperature change}} + Q_{text{vaporization}}$

Calculating the value:
$Q_{text{total}} approx 17450 , text{J} + 113600 , text{J} approx 131050 , text{J}$

Hence, approximately 131050 J of energy is required to raise the temperature of 50 grams of water from 25°C to its boiling point and vaporize it completely.

Example 3: Determining How Much Energy is Required to Vaporize

Let’s say we have a sample of ethanol with a mass of 250 grams. We want to determine how much energy is required to completely vaporize the ethanol, given that its heat of vaporization is 38.56 kJ/mol.

Given:
– Mass of ethanol (m): 250 grams
– Heat of vaporization of ethanol $( Delta H_v$ ): 38.56 kJ/mol / 46.07 g/mol

$Q = m cdot Delta H_v = 250 , text{g} times left(frac{38.56 , text{kJ/mol}}{46.07 , text{g/mol}}right)$

Calculating the value:

$Q approx 205.8 , text{kJ}$

Therefore, approximately 205.8 kJ of energy is required to completely vaporize 250 grams of ethanol.

Understanding the energy required for vaporization is essential in various scientific and engineering applications. By applying the formula and concepts discussed in this article, you can calculate the energy required to vaporize a substance accurately. Whether you are studying thermodynamics, chemistry, or simply curious about the physics behind phase transitions, mastering the calculation of energy required for vaporization will provide you with valuable insights into the behavior of substances at different temperatures and pressures.

Numerical Problems on How to find energy required to vaporize

Problem 1:

A substance requires 2000 J of energy to vaporize 1 gram of it. If 5 grams of the substance is vaporized, calculate the total energy required.

Solution:

Given:
Energy required to vaporize 1 gram of the substance = 2000 J
Mass of the substance vaporized = 5 grams

To find the total energy required, we can use the formula:

$text{Total energy required} = text{Energy required to vaporize 1 gram} times text{Mass of the substance vaporized}$

Substituting the given values,

$text{Total energy required} = 2000 , text{J/g} times 5 , text{g}$

$text{Total energy required} = 10000 , text{J}$

Therefore, the total energy required to vaporize 5 grams of the substance is 10000 J.

Problem 2:

A certain liquid requires 1500 J of energy to vaporize 100 grams of it. Calculate the energy required to vaporize 2.5 kg of the liquid.

Solution:

Given:
Energy required to vaporize 100 grams of the liquid = 1500 J
Mass of the liquid to be vaporized = 2.5 kg

To find the energy required to vaporize 2.5 kg of the liquid, we can first convert the mass from kg to grams:

$2.5 , text{kg} = 2.5 times 1000 , text{g} = 2500 , text{g}$

Using the formula:

$text{Total energy required} = text{Energy required to vaporize 100 grams} times text{Mass of the liquid to be vaporized}$

Substituting the given values,

$text{Total energy required} = 1500 , text{J/g} times 2500 , text{g}$

$text{Total energy required} = 3750000 , text{J}$

Therefore, the energy required to vaporize 2.5 kg of the liquid is 3750000 J.

Problem 3:

How to find energy required to vaporize — Image by WilfriedC – Wikimedia Commons, Wikimedia Commons, Licensed under CC BY-SA 3.0.

It takes 4800 J of energy to vaporize 200 grams of a substance. How much energy is required to vaporize 3 kg of the same substance?

Solution:

Given:
Energy required to vaporize 200 grams of the substance = 4800 J
Mass of the substance to be vaporized = 3 kg

To find the energy required to vaporize 3 kg of the substance, we can first convert the mass from kg to grams:

$3 , text{kg} = 3 times 1000 , text{g} = 3000 , text{g}$

Using the formula:

$text{Total energy required} = text{Energy required to vaporize 200 grams} times text{Mass of the substance to be vaporized}$

Substituting the given values,

$text{Total energy required} = 4800 , text{J/g} times 3000 , text{g}$

$text{Total energy required} = 14400000 , text{J}$

Therefore, the energy required to vaporize 3 kg of the substance is 14400000 J.

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