The total energy required to heat 38.0 g of liquid water from 55 °C to 100 °C and change it to steam at 100 °C is 94.4 kJ.
First, let's break the problem into two parts:
1. Heating the liquid water from 55 °C to 100 °C
2. Changing the liquid water to steam at 100 °C
For part 1, we can use the formula:
Q = m * c * ΔT
Where Q is the amount of heat energy required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Plugging in the values we have:
Q = 38.0 g * 4.18 J/goC * (100 °C - 55 °C)
Q = 8,692.4 J
This tells us that it takes 8,692.4 J of energy to heat the water from 55 °C to 100 °C.
For part 2, we need to find the energy required to change the water to steam. This is known as the molar heat of vaporization, which is the amount of energy required to turn one mole of a substance from a liquid to a gas.
The molar heat of vaporization of water is 40.6 kJ/mol. We need to figure out how many moles of water we have so we can use this value.
To do this, we can use the molar mass of water, which is approximately 18 g/mol.
38.0 g / 18 g/mol = 2.11 mol
So we have 2.11 moles of water.
Now we can use the formula:
Q = n * ΔH
Where Q is the amount of energy required, n is the number of moles of water, and ΔH is the molar heat of vaporization.
Plugging in the values we have:
Q = 2.11 mol * 40.6 kJ/mol
Q = 85.7 kJ
This tells us that it takes 85.7 kJ of energy to change 38.0 g of water to steam at 100 °C.
To find the total energy required, we can add the energy required for part 1 and part 2:
Total energy = 8,692.4 J + 85.7 kJ
Total energy = 94.4 kJ