Calculate ΔSsys, ΔSsurr, and ΔStot when 245 g of ice at 0 oC is added to 32 g of steam at 115 oC in a poorly insulated container. During this addition, 4530 J of heat energy is lost to the surroundings. The values of Cp,m for water, ice, and steam are 75.3 J/K mol, 37.7 J/K mol, and 35.5 J/K mol respectively. The ΔHfus for water is 6.02 kJ/mol and the ΔHvap for water is 40.656 kJ/mol.
Given data:
The mass of ice = 245 g = 245/18 = 13.61 mol, its temperature (T1) = 0 oC = 273 K
The mass of steam = 32 g = 32/18 = 1.78 mol, its temperature (T2) = 115 oC = 115 + 273 = 388 K
i.e. ΔT = 388 - 273 = 115 K
The molar specific heat capacity of water (Cp,m for water) = 75.3 J/K mol
The molar specific heat capacity of ice (Cp,m for ice) = 37.7 J/K mol
The molar specific heat capacity of steam (Cp,m for steam) = 35.5 J/K mol
ΔHfus for water = 6.02 kJ/mol
ΔHvap for water = 40.656 kJ/mol.
The heat released during the given addition process = 4530 J
ΔSsys = Qsys / ΔT
= 4530/115 K
= 39.4 J/K
ΔSsurr = Qsurr / ΔT
= -4530/115 K
= -39.4 J/K
ΔStot = ΔSsys + ΔSsurr
= 39.4 - 39.4
= 0
Calculate ΔSsys, ΔSsurr, and ΔStot when 245 g of ice at 0 oC is added to...
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