Find the mass of liquid water required to absorb 4.81×104 kJ of heat energy on boiling.
Answer in units of g.
What is the initial temperature of the water? If we assume the water is at the boiling point (100 C), look up the heat of vaporization.
q = mass H2O x heat vaporization
Look up the density of water @ 100 C, and convert mass to volume by
mass = volume x density
Post your work if you get stuck.
Well, the answer is actually quite simple. You see, when water boils, it undergoes a phase change and turns into steam. And do you know what they say about steam? It's like water, but with a lot more enthusiasm! So, to calculate the mass of liquid water required to absorb all that heat energy, we just need a little bit of chemistry and a whole lot of imagination.
The enthalpy of vaporization, which is the amount of heat energy required to convert a substance from a liquid to a gas at its boiling point, for water is about 40.7 kJ/mol. Now, in order to find the molar amount of water we need to boil, we can simply divide the total heat energy (4.81×104 kJ) by the enthalpy of vaporization (40.7 kJ/mol).
So, 4.81×104 kJ divided by 40.7 kJ/mol gives us approximately 1182.87 mol of water. Now, we just need to convert this to grams. The molar mass of water is approximately 18.02 g/mol.
Multiplying 1182.87 mol by 18.02 g/mol gives us the mass of liquid water required, which is approximately 21,326.69 g.
So, the mass of liquid water required to absorb 4.81×104 kJ of heat energy on boiling is approximately 21,326.69 g. That's a whole lot of water!
To find the mass of liquid water required to absorb 4.81×10^4 kJ of heat energy on boiling, we need to use the specific heat capacity and the heat of vaporization of water.
1. Calculate the heat required to increase the temperature of water from its boiling point (100°C) to its boiling point at the given pressure:
Heat = Mass × Specific Heat Capacity × Temperature Change
The specific heat capacity of water is 4.186 J/g°C.
The temperature change is (100°C - 100°C) = 0°C.
So, Heat = Mass × 4.186 J/g°C × 0°C
2. Calculate the heat required for the phase change from liquid to vapor (boiling):
Heat = Mass × Heat of Vaporization
The heat of vaporization for water is 2260 J/g.
3. Add the two heats together to get the total heat required:
Total Heat = Heat for temperature change + Heat for phase change
4. Now, we can calculate the mass of water:
Total Heat = Mass × 4.186 J/g°C × 0°C + Mass × 2260 J/g
Mass × (4.186 J/g°C + 2260 J/g) = 4.81×10^4 kJ × 1000
5. Convert kJ to J and solve for Mass:
Mass = (4.81×10^4 kJ × 1000) / (4.186 J/g°C + 2260 J/g)
Solving this equation will give us the mass of liquid water required to absorb 4.81×10^4 kJ of heat energy on boiling.
To find the mass of liquid water required to absorb 4.81×104 kJ of heat energy on boiling, we need to use the specific heat capacity and the heat of vaporization of water.
The specific heat capacity of water is approximately 4.18 J/g°C, meaning it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
The heat of vaporization of water is approximately 40.7 kJ/mol, meaning it takes 40.7 kilojoules of energy to convert 1 mole of liquid water into steam at boiling point.
First, let's calculate the heat energy required to raise the temperature of water from its boiling point (100°C) to boiling point.
The specific heat capacity equation is:
Q = m * c * ΔT
where:
Q = heat energy in Joules
m = mass of the substance in grams
c = specific heat capacity in J/g°C
ΔT = change in temperature in °C
Since we need to raise the temperature from boiling point to boiling point, ΔT will be 0. Therefore, the first part of the heat energy required will be 0.
Next, we need to calculate the heat energy required for the phase change from liquid to gas (vaporization). The equation for this is:
Q = m * ΔHvap
where:
Q = heat energy in Joules
m = mass of the substance in grams
ΔHvap = heat of vaporization in J/g
Now we can plug in the values:
Q = 4.81×104 kJ = 4.81×107 J (since 1 kJ = 1000 J)
c = 4.18 J/g°C
ΔHvap = 40.7 kJ/mol = 40.7×103 J/g
For the first part of the equation, we have m * 4.18 * 0 = 0 J.
For the second part of the equation, we have m * 40.7×103 J/g.
So, the equation becomes:
4.81×107 J = m * 40.7×103 J/g
Now we can solve for m:
m = 4.81×107 J / 40.7×103 J/g
m = 1183.56 g
Therefore, the mass of liquid water required to absorb 4.81×104 kJ of heat energy on boiling is approximately 1183.56 grams.