Determine the entropy change, in J/K, for melting 26.0 g of solid water at 273 K. For water, H = 6005 .
∆S=q/T
q=mol×∆H fus
Mol=mass/molar mass=26gm/(18gm/mol)=1.44mol
∆H fus=6005J/K
Then q=1.44mol×6005J/mol=8647.2J
Therefor∆S=8647.2J/273K=31.67J/K
Determine the entropy change, in J/K, for melting 26.0 g of solid water at 273 K....
The change in enthalpy when 1 mol of ice is melted at 273K is 6008 J Heat Capacity of liquid water, Cp_{L} = 75.44 J/mol K Heat Capacity of solid water, Cp_{S} = 38J/mol K Enthalpy chage of melting at 273K, \Delta H_{273}=6008 J Calculate the standard enthalpy of fusion for ice. Calculate the heat released when 100 g of water supercooled at 250K solidify Initial T=25°C=298K Thanks
i got this question wrong. the book previously said that the
units of entropy are J/(mol*K). you can get mol with the grams that
they gave, so why not use it?
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