Calculate the heat in kilojoules needed to vaporize 8.00kg of water at 100∘C.
To calculate the heat needed to vaporize 8.00 kg of water at 100∘C, you need to use the formula:
Heat (Q) = mass (m) × specific heat capacity (c) × change in temperature (ΔT)
However, this formula only calculates the heat required to raise the temperature of the water, and not to vaporize it. To include the heat required for vaporization, you need to add the product of the mass of the water and the heat of vaporization (Hvap).
The heat equation including both components becomes:
Q = mcΔT + mHvap
First, let's calculate the heat required to raise the temperature of the water from 100∘C to its boiling point, which is 100∘C:
Q1 = mcΔT
= (8.00 kg) × (4.18 kJ/kg⋅∘C) × (100∘C - 100∘C)
= 0 kJ
Since the temperature does not change, there is no heat required to increase the temperature.
Next, let's calculate the heat of vaporization (Hvap). The heat of vaporization of water is approximately 40.7 kJ/mol.
To find the heat of vaporization in kJ for 8.00 kg of water, we need to convert the mass from kg to moles. To do this, we will use the molar mass of water, which is approximately 18.02 g/mol:
moles of water (n) = mass of water / molar mass of water
= (8.00 kg) / (18.02 g/mol)
= 444.83 mol
Now, we can calculate the heat of vaporization:
Hvap = Q2 / n
= (40.7 kJ/mol) / (444.83 mol)
= 0.0915 kJ/mol
Finally, let's calculate the total heat required to vaporize 8.00 kg of water:
Q = mcΔT + mHvap
= 0 kJ + (8.00 kg) × (0.0915 kJ/mol)
= 0.732 kJ
Therefore, the heat required to vaporize 8.00 kg of water at 100∘C is 0.732 kilojoules.
To calculate the heat required to vaporize water at 100℃, we need to consider two steps:
1. Heating the water from 0℃ to 100℃
2. Vaporizing the water at 100℃
Step 1: Heating the water from 0℃ to 100℃
The specific heat capacity of water is approximately 4.18 J/g℃.
Formula:
Q1 = m * c * ΔT
where:
Q1 = heat required to heat the water
m = mass of the water
c = specific heat capacity of water
ΔT = change in temperature
Given:
m = 8.00 kg
c = 4.18 J/g℃
ΔT = (100℃ - 0℃) = 100℃
Converting the mass of the water from kg to grams:
m = 8.00 kg × 1000 g/kg = 8000 g
Calculating Q1:
Q1 = (8000 g) * (4.18 J/g℃) * (100℃)
Q1 = 3,344,000 J
Step 2: Vaporizing the water at 100℃
The heat of vaporization of water is approximately 2260 J/g.
Formula:
Q2 = m * H
where:
Q2 = heat required to vaporize the water
m = mass of the water
H = heat of vaporization of water
Given:
m = 8.00 kg
H = 2260 J/g
Converting the mass of the water from kg to grams:
m = 8.00 kg × 1000 g/kg = 8000 g
Calculating Q2:
Q2 = (8000 g) * (2260 J/g)
Q2 = 18,080,000 J
Total heat required:
Total heat = Q1 + Q2
Total heat = 3,344,000 J + 18,080,000 J
Total heat = 21,424,000 J
Converting the total heat from joules to kilojoules:
Total heat in kilojoules = 21,424,000 J / 1000
Total heat in kilojoules ≈ 21,424 kJ
Therefore, the heat required to vaporize 8.00 kg of water at 100℃ is approximately 21,424 kilojoules.
q = mass(grams) x Hvaporization (J/g)
This will give q in J/g, then change J to kJ.