Question

At room temperature, it takes approximately 2.45 × 10³ J to evaporate 1.00 g of water.

A. Specific heat capacity of water
B. Heat of fusion of water
C. Heat of vaporization of water
D. Thermal conductivity of water

Answer 1

The quantity 2,250 J per gram of water refers to the heat of vaporization, which is the energy needed to evaporate water. This is different from the specific heat capacity of water, which is 4.184 J/g °C. Calculating the evaporation time involves knowing both the initial mass of the water and the rate of energy input.

Step-by-step explanation:

The information about the amount of energy required to evaporate water relates to the heat of vaporization of water. This is a thermodynamic property that indicates the amount of heat needed to turn 1 gram of a liquid into a gas at constant temperature and pressure.

In this case, the heat of vaporization of water is about 2,250 J per gram. This is distinct from the specific heat capacity, which is the energy required to raise the temperature of 1 gram of water by 1 °C. For water, this is 4.184 J/g °C.

Given it takes 2,250 J to evaporate 1 gram of water, if you're injecting energy at a constant rate throughout the process, calculating the time to completely evaporate the water would require knowing the mass of the water initially present and the rate at which energy is being added.

The process involves first heating the water to boiling, then supplying additional energy to change the state from liquid to vapor without changing the temperature, represented by the heat of vaporization. The entire amount of energy needed would be the sum of energy to heat the water to boiling plus the energy represented by the heat of vaporization.

Answer 2

The heat of vaporization of water is approximately 2,250 J per gram. The specific heat capacity of water is 4.184 J/g °C. The correct answer is (C) Heat of vaporization of water.

Step-by-step explanation:

The heat of vaporization of water is approximately 2,250 J per gram, meaning that every gram of water that changes from liquid to gas (vapor) requires an energy input of 2,250 J.

The specific heat capacity of water is 4.184 J/g °C. This means that to heat 1 gram of water by 1 °C, it requires 4.184 J of energy.

The correct answer to the question is C. Heat of vaporization of water.

The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol. The vaporization is the opposite process of condensation. The heat of condensation is defined as the heat released when one mole of the substance condenses at its boiling point under standard pressure.

The specific latent heat of a substance is the quantity of heat energy required to change the state of a unit mass of a substance. EL=ml where EL is the heat transferred, in joules, m is the mass, in kilograms, and l is the latent heat in joules per kilogram. The SI unit for specific latent heat is Jkg.