a. Heat needed to melt 1 mol of ice to liquid water at 273 K
b. 6.02 KJ/mol
c. 1 mol of water = 18 g
Heat of fusion = molar heat of fusion / molar mass = 6.02 / 18 = 0.334 KJ/g
d. Heat of fusion in J / g = KJ / g * 1000 = 334 J / g
e. 1 cal = 4.184 J
Heat of fusion in cal / g = heat of fusion in J/g / 4.184 = 79.83 cal / g
f. 40.7 KJ / mol
g. 40.7 KJ/ mol
h. Heat of vaporisation in KJ / g = 40.7 / 18 = 2.26 KJ / g
Define each of the following: The molar heat of fusion of water a. met Imol of...
Calculate the heat required in Joules to convert 18.0 grams of water ice at a temperature of -20° C to liquid water at the normal boiling point of water. Given: -specific heat of ice = 2.09 J/g°C -specific heat of liquid water = 4.184 J/g°C -specific heat of water vapor = 2.03 J/g°C -molar heat of fusion of water = 6.02 kJ/mol -molar heat of vaporization of water = 40.7 kJ/mol
11. The following would be required for calculations of heat flow in which of the heating curve steps ? Molar heat of vaporization of water (AH vap = 40.7 kJ/mol) Specific heat of ice (Cice = 2.09 J/g °C) Molar heat of fusion of water (AH fus = 6.02 kJ/mol) Specific heat of water (C H20 = 4.18 J/g °C) Specific heat of steam(C steam = 2.01 J/g °C) Heating Curve for Water Degrees Celsius -50+ 0 400 800 1200...
Determination Of The Molar Heat of Fusion Of Ice Purpose The purpose of this result with its reported value that is s 6.01 kJ/mol. The molar heat of fusion (AH) of ice is the amount of heat required to melt 1 mole of ice experiment is to determine the molar heat of fusion of ice and compare the experimental Definition Heat of fusion of ice is the amount of heat required to melt ice. The more the ice to melt,...
Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol⋅K and that the heat capacity of H2O(s) is 37.7 J/mol⋅K, calculate the heat of fusion of water at -13 ∘C.
Given that the heat of fusion of water is 6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J>mol · K, and that the heat capacity of H2O(s) is 37.7 J/mol · K, calculate the heat of fusion of water at – 10 °C.
Given that the heat of fusion of water is +6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol⋅K and that the heat capacity of H2O(s) is 37.7 J/mol⋅K, calculate the heat of fusion of water at -14 ∘C.
Please answer clearly & correctly. Selected properties of water Heat of fusion at the normal melting point: The heat of vaporization at the normal boiling point: Liquid water: 333.55 J/g 2257 J/g Cp = 1.00 cal/(g. K) = 4.184 J/(g. K) or 75.33 J/(mole. K) ρ= 1 .000 g/cm3 Cp = 2.05 J/(g. K) or 38, l J/(mole K) ρ=0.917 g/cm3 Water ice at 0°C and 1 atm Heat capacity of water vapor, H20(g) Cvm 28.03 J/(mol-K).
Given that the heat of fusion of water is +6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 J/mol⋅K and that the heat capacity of H2O(s) is 37.7 J/mol⋅K, calculate the heat of fusion of water at -15 ∘C. Express your answer using two significant figures. deltaH= ___?____ kJ/mol
water has a molar heat capacity of 75.38 J/(mol 12. Water has a molar heat capacity of 75.38 J/(mol x °C) and its vaporization enthalpy at 100°C is 40.7 kJ/mol. How much energy is needed to convert 36 g liquid H20 at 70.0°C to steam at 100°C? A) 85.9 kJ B) 126 kJ C) 77.6 kJ D) 81.4 kJ E) 45.2 kJ
how much heat is released when 10.0 g of steam (water vapor ) at 105.0 C is cooled to liquid water at 25 C? S(water) = 4.18 J/g.C. ... S(steam) = 2.01 j/ g.C the heat of fusion of water is 6.02 KJ/ mol. The heat of vaporization of water is 40.7 KJ/mol