Completed question:
A sample of steam with a mass of 0.535 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C)
Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?
Answer:
73.9°C
Explanation:
The steam is already at the water boiling point (100°C at 1 atm), so it will first lose heat to be condensed, by the equation:
Q1 = -n*ΔH∘vap*1000
Where n is the number of moles, ΔH∘vap is the enthalpy of evaporation, and the minus sign indicates that the heat is being lost. The equation is multiplied by 1000 because ΔH∘vap is in kJ, and the result must be in J.
Then, when it is in a liquid state, it will change heat with the cold water presented. The hot one will lose heat (Q2) and the cold one will gain heat (Q3) as the equation:
Q2 = mh*c*ΔTh
Q3 = mc*c*ΔTc
Where mh is the mass of the hot water, ΔTh is the variation of the temperature of the hot water (final - initial), mc is the mass of the cold water, and ΔTc is the variation of the temperature of the cold water.
At equilibrium they both will have the same final temperature (T), and, because the system doesn't change heat with the surroundings, the sum of all those heats must be 0.
n = mass/molar mass
Molar mass of water: 18g/mol
n = 0.535/18 = 0.0297 mol
Q1 + Q2 + Q3 = 0
-0.0297*40.7*1000 + 0.535*4.18*(T - 100) + 4.40*4.18*(T - 5.0) = 0
-1208.79 + 2.2363T - 223.63 + 18.392T - 91.96 = 0
20.6283T = 1524.38
T = 73.9°C
The final temperature of the mixture is mathematically given as
T = 73.9°C
Question Parameter(s):
Generally, the equation for the heat is mathematically given as
Q1 = -n*dH∘vap*1000
Q2 = mh*c*dTh
Q3 = mc*c*dTc
Therefore
Q1 + Q2 + Q3 = 0
-0.0297*40.7*1000 + 0.535*4.18*(T - 100) + 4.40*4.18*(T - 5.0) = 0
20.6283T = 1524.38
T = 73.9°C
In conclusion, The temperature is
T = 73.9°C
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Complete question:
A sample of steam with a mass of 0.535 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?
