calculate the amount of heat (in kj) required to convert 140.1 g of water to steam at 100c
To calculate the amount of heat required to convert water to steam, we need to consider two components: the heat required to reach the boiling point (100°C) and the heat required to actually convert the water to steam.
First, let's calculate the heat required to raise the temperature of the water from its initial temperature to the boiling point (100°C). We can use the specific heat capacity of water to do this calculation. The specific heat capacity of water is 4.18 J/g°C.
Formula: Q = m * c * ΔT
Where:
Q = Heat energy (in joules)
m = Mass of the substance (in grams)
c = Specific heat capacity of the substance (in J/g°C)
ΔT = Change in temperature (in °C)
Inputting the values into the formula:
Q = 140.1 g * 4.18 J/g°C * (100°C - 25°C)
Calculating ΔT:
Q = 140.1 g * 4.18 J/g°C * 75°C
Now we have the amount of heat required (Q) in joules. To convert this to kilojoules (kJ), we divide by 1000.
Converting to kJ:
Q = 140.1 g * 4.18 J/g°C * 75°C / 1000
Now, we'll calculate the heat required to convert the water to steam. This is known as the heat of vaporization. The heat of vaporization for water is 40.7 kJ/mol.
To convert the mass of water to moles, we need to divide by the molar mass of water.
The molar mass of water (H2O) is:
H: 1.01 g/mol
O: 16.00 g/mol
Adding them up:
1.01 g/mol + 2 * 16.00 g/mol = 18.02 g/mol
Now, let's calculate the moles of water:
moles of water = 140.1 g / 18.02 g/mol
Finally, we can calculate the heat required to convert water to steam:
heat required = moles of water * heat of vaporization
Calculating:
heat required = (140.1 g / 18.02 g/mol) * 40.7 kJ/mol
In this case, we didn't convert to kJ, as the heat of vaporization is already given in kJ/mol.
Now, you can add the two calculated amounts of heat (in kJ) to find the total amount of heat required to convert 140.1 g of water to steam at 100°C.