The heat required to vaporize 33.8 g of water at is .
Further explanation:
Properties are categorized into two types:
1. Intensive properties:
These properties depend on the nature of the substance and not on the size of the system. If the system is further divided into a number of subsystems, the values of intensive properties remain unchanged. Temperature, refractive index, molarity, concentration, pressure, and density are some of the examples of intensive properties.
2. Extensive properties:
These properties depend on the amount of the substance. These are additive in nature if a single system is divided into several subsystems. Mass, heat, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.
Heat of vaporization:
It is the amount of heat that is needed to convert a unit mass of the liquid into its vapors. It is also known as the enthalpy of vaporization or heat of evaporation. It is represented by . Its unit is kJ/mol or J/mol.
The formula to calculate the moles of water is as follows:
…… (1)
The given mass of water is 33.8 g.
The molar mass of water is 18.01 g/mol.
Substitute these values in equation (1).
The heat of vaporization of water at is 40.7 kJ/mol. This implies 40.7 kJ of energy is required to convert 1 mole of water into vapors. So the heat required to convert 1.876 moles of water is calculated as follows:
Therefore the heat required to vaporize 33.8 g of water at is 76.3532 kJ.
Learn more:
1. Calculate the enthalpy change using Hess’s Law: <u>brainly.com/question/11293201
</u>
2. Find the enthalpy of decomposition of 1 mole of MgO: <u>brainly.com/question/2416245
</u>
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: heat of vaporization, enthalpy of vaporization, heat of evaporation, 33.8 g, water, 40.7 kJ/mol, 76.3532 kJ, heat, intensive, extensive, molar mass of water, 18.01 g/mol, 1.876 mol.