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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 -74°C.

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Beverley Smith
Beverley SmithLv2
28 Sep 2019

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Related questions

You have a 1.00-mole sample of water at -30 oC and you heat it until you have gaseous water at 140 oC. What is the energy absorbed by water in the whole process? Use the following data.

 

Specific heat capacity of ice = 2.03 J oC-1 g-1

Specific heat capacity of water = 4.18 J oC-1 g-1

Specific heat capacity of steam = 2.02 J oC-1 g-1

Heat of fusion () = 6.02 kJ mol-1 (at 0 oC)

Heat of vaporization () = 40.7 kJ mol-1 (at 100 oC)

How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0 ∘C, to steam at 109.0 ∘C? The heat capacity of ice is 2.09 J/g⋠∘C and that of steam is 2.01 J/g⋠∘C, the heat of fusion for water is 6.02 kJ/mol, and the heat of vaporization for water is 40.7 kJ/mol. ( the answer needs to be in kJ) :)

Calculate the enthalpy for the freezing of one mole of water at 1.00 bar and -10.0˚C. The heat capacities for ice and liquid water are 36.9 J/mol·K and 74.5 J/mol·K, respectively (assume that these values are independent of temperature.) The enthalpy for the freezing of water at 0.0˚C is -6.025 kJ/mol.

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