65.0 mL of water is heated to its boiling point. How much heat in Kj is required to vaporize it? (Assume a density of 1.00 g/mL)
use dimensional analysis and the heat vaporization table.
Water = 40.7 kj/mol.
Start with the information that you were given and make sure that all of your units cancel out!
(65.0mL)(1.00g/1mL)(1 mole H2O/18.0g H2O)(40.7kj/1 mole H2O) = 146.9 kj.
There are 3 significant figures, so 147kj is your answer.
This doesn't help me.
Can someone please answer this.
Well, well, well, looks like we're in for some hot water! Let's calculate the amount of heat required to vaporize that liquid comedy gold.
First, we need to find the mass of water. Given that the density of water is 1.00 g/mL, and we have 65.0 mL, we can conclude that we have 65.0 grams of water. I must say, it's always refreshing to have a liquid with such a perfectly balanced density!
Now comes the fun part, vaporizing. To get the heat required, we need to find the specific heat capacity of water. On a serious note, water has a specific heat capacity of about 4.18 J/g·°C.
Since we're working with kilojoules, let's convert our answer. There are 1,000 J in 1 KJ, so we get:
Heat = (mass of water) x (specific heat capacity) x (temperature change)
But wait, the temperature change! We're heating it to its boiling point, so we need to find the temperature change from room temperature (let's say 25°C) to boiling point (100°C). That gives us a ΔT of 75°C.
Plug in the numbers and we get:
Heat = (65.0 g) x (4.18 J/g·°C) x (75°C) ÷ (1,000 J/KJ)
So, if my calculations and comedy don't evaporate into thin air, you'd need approximately 202.725 KJ of heat to vaporize that 65.0 mL of water. That's enough heat to make any clown's argument hold water!
To calculate the heat required to vaporize water, we need to know the heat of vaporization for water, which is the amount of energy needed to change a substance from a liquid to a gas at its boiling point.
The heat of vaporization for water is approximately 40.7 kJ/mol. To use this value, we need to convert the given volume of water to moles. We can do this using the density of water, which is given as 1.00 g/mL:
65.0 mL of water * 1.00 g/mL = 65.0 g of water
Next, we can convert grams to moles using the molar mass of water, which is approximately 18.0 g/mol:
65.0 g of water / 18.0 g/mol ≈ 3.61 mol of water
Now that we have the number of moles, we can calculate the heat required to vaporize the water:
Heat = moles of water * heat of vaporization
Heat = 3.61 mol * 40.7 kJ/mol
Heat ≈ 146.8 kJ
Therefore, approximately 146.8 kJ of heat is required to vaporize 65.0 mL of water.
q = mass x heat vaporization
Then tell me what you don't understand. All you need to do is to substitute the mass of 65.0 mL water (65.0 grams) and surely you have a table giving the heat vaporization of water. That calculates q, the heat required to vaporize water at it's boiling point. How did I get the mass of 65.0 grams?
mass = density x volume
mass = 1.00 g/mL x 65.0 mL = 65.0 grams.