Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 kJ/mol*K and that the heat capacity of H2O(s) is 37.7 kJ/mol*K, calculate the heat of fusion of water at -11 K.
A substance's heat of fusion is the quantity of thermal energy involved in the solid/liquid phase change. When water freezes it releases 6.02kJ for each 18.0 grams (1 mole) that freezes. When it melts, each mole must gain 6.02 kJ of energy. Remember that there is no temperature change during phase changes - this is a change that, for pure water, occurs at 0 oC! Asking what water's heat of fusion is at -12 oC is kind of strange.....the heat of fusion of pure water is always the same but it's only important or is only involved when the substance is freezing or melting which doesn't occur at -12 oC.
But I guess for general learning purposes, this would be how to solve it.
The water is supercooled.
so it has a deficit of:
11° x (75.2 J/mol·K - 37.7 J/mol·K )
= 562.5 J/mol.
therefore you'll see
-6.02 kJ/mol + 0.562
= -5.458 kJ/mol
Why does the temperature does not change ? Shouldn't it in Kelvin because the heat capacity is J/mol K ?
Well, it seems like you're trying to freeze water at -11K. That's pretty chilly! But don't worry, I'm here to help you with this frosty situation.
To calculate the heat of fusion of water at -11K, we can use the equation:
ΔH_fusion = ΔH_solid + ΔH_liquid
Now, let's start by calculating ΔH_solid, the enthalpy change of water from ice to water at -11K.
ΔH_solid = heat capacity of H2O(s) * change in temperature
To find the change in temperature, we subtract the final temperature (-11K) from the freezing point of water (0K):
Change in temperature = -11K - 0K = -11K
So, ΔH_solid = 37.7 kJ/mol*K * (-11K) = -414.7 kJ/mol
Now, let's calculate ΔH_liquid, the enthalpy change of water from solid to liquid at -11K.
ΔH_liquid = heat of fusion of water + heat capacity of H2O(l) * change in temperature
Again, let's find the change in temperature, which is just the difference between the final and initial temperatures:
Change in temperature = -11K - 0K = -11K
ΔH_liquid = -6.02 kJ/mol + 75.2 kJ/mol*K * (-11K) = -6.02 kJ/mol + (-827.2 kJ/mol) = -833.22 kJ/mol
Finally, we can calculate the heat of fusion of water at -11K by adding ΔH_solid and ΔH_liquid:
ΔH_fusion = ΔH_solid + ΔH_liquid = -414.7 kJ/mol + (-833.22 kJ/mol) = -1247.92 kJ/mol
So, the heat of fusion of water at -11K is approximately -1247.92 kJ/mol.
Stay cool, my friend!
To calculate the heat of fusion of water at -11 K, we can use the formula:
ΔH = ΔHfus + ΔHsolidification
where:
ΔH is the total heat change
ΔHfus is the heat of fusion
ΔHsolidification is the heat change during solidification
Given that the heat capacity of H2O(l) is 75.2 kJ/mol*K and the heat capacity of H2O(s) is 37.7 kJ/mol*K, we can determine the heat change during solidification using the formula:
ΔHsolidification = n * ΔCsolid * ΔT
where:
n is the number of moles of water
ΔCsolid is the heat capacity of H2O(s)
ΔT is the change in temperature
To calculate the number of moles of water, we can use the formula:
n = mass / molar mass
Now let's calculate each step:
Step 1: Calculate the number of moles of water
Given that we have 1 mol of water, the number of moles is 1.
Step 2: Calculate the heat change during solidification
ΔHsolidification = 1 mol * (37.7 kJ/mol*K) * (-11 K)
ΔHsolidification = -414.7 kJ
Step 3: Calculate the total heat change
Since we want to find the heat of fusion at -11 K, the solidification process does not occur. Therefore, ΔHsolidification is zero.
ΔH = ΔHfus + ΔHsolidification
ΔH = ΔHfus + 0
ΔH = ΔHfus
So, the heat of fusion of water at -11 K is -414.7 kJ.