L>jiyuushikan.org: Molar Heat of FusionMolar Heat of FusionReturn to the Time-Temperature Graph fileRerevolve to Thermochemisattempt MenuHere is the definition of the molar heat of fusion:the amount of warm important to melt (or freeze) 1.00 mole of a substance at its melting pointKeep in mind the 2 vital factors:1) It"s 1.00 mole of a substance2) tbelow is no temperature changeKeep in mind the reality that this is an extremely particular worth. It is just for one mole of substance melting. The molar warmth of fusion is a critical component of power calculations because it tells you just how much energy is needed to melt each mole of substance on hand also. (Or, if you are cooling off a substance, exactly how much power per mole to remove from a substance as it solidifies.Eexceptionally substance has actually its own molar warm of fusion.The systems for the molar warm of fusion are kilojoules per mole (kJ/mol). Sometimes, the unit J/g is used. In that case, the term warmth of fusion is used, via the word "molar" being removed. See Example #3 listed below.The molar warm of fusion for water is 6.02 kJ/mol. As you go about the Internet, you will certainly watch other values used. For instance, 6.01 is a popular value and you periodically check out 6.008. I flourished up via 6.02, so I"ll stick to it.Molar warmth values can be looked up in referral publications.The molar warmth of fusion equation looks prefer this:q = ΔHfus (mass/molar mass)The meanings are as follows:1) q is the full amount of heat involved2) ΔHfus is the symbol for the molar warm of fusion. This value is a constant for a given substance.3) (mass/molar mass) is the department to gain the number of moles of substanceExample #1: 31.5 g of H2O is being melted at its melting suggest of 0 °C. How many kJ is required?Solution:plug the correct worths into the molar heat equation presented aboveq = (6.02 kJ/mol) (31.5 g / 18.0 g/mol)Example #2: 53.1 g of H2O exists as a liquid at 0 °C. How many kJ should be rerelocated to rotate the water right into a solid at 0 °CSolution:note that the water is being frozen and also that tright here is NO temperature change. The molar warm of fusion value is provided at the solid-liquid phase readjust, REGARDLESS of the direction (melting or freezing).q = (6.02 kJ / mol) (53.1 g / 18.0 g/mol)Example #3: Calculate the heat of fusion for water in J/gSolution:divide the molar warm of fusion (expressed in Joules) by the mass of one mole of water.(6020 J / mol) / (18.015 g/mol)This value, 334.166 J/g, is referred to as the warmth of fusion, it is not called the molar warm of fusion. When this value is provided in difficulties, the 334 J/g worth is what is most-often used.Example #4: Using the warm of fusion for water in J/g, calculate the power needed to melt 50.0 g of water at its melting suggest of 0 °C.Solution:multiply the warm of fusion (expressed in J/g) by the mass of the water connected.(334.166 J/g) (50.0 g) = 16708.3 J = 16.7 kJ (to 3 sig figs)Example #5: By what aspect is the energy need to evapoprice 75 g of water at 100 °C greater than the energy required to melt 75 g of ice at 0 °C? Solution:Notice how the amounts of water are the same.
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This is delibeprice. The etop quality is necessary, not the amount.Change the 75 g to one mole and solve:40.7 kJ / 6.02 kJ = 6.76Change the amount to 1 gram of water and also solve:2259.23 J / 334.166 J = 6.76If you insisted that you have to do it for 75 g, then we have this:(75 g * 2259.23 J/g) / (75 g * 334.166 J/g) = ???You can view that the 75 cancels out, leaving 6.76 for the answer.Return to the Time-Temperature Graph fileRerotate to Thermochemistry Menu