one mole of ice at 0 C is put into contact with one mole of steam at 100 C in an
ID: 890256 • Letter: O
Question
one mole of ice at 0 C is put into contact with one mole of steam at 100 C in an adiabatic enclosure. The pressure is held constant at one atmosphere. What is the final composition and temperature of the system? one mole of ice at 0 C is put into contact with one mole of steam at 100 C in an adiabatic enclosure. The pressure is held constant at one atmosphere. What is the final composition and temperature of the system? one mole of ice at 0 C is put into contact with one mole of steam at 100 C in an adiabatic enclosure. The pressure is held constant at one atmosphere. What is the final composition and temperature of the system?Explanation / Answer
To calculate the final temperature we will look at each transition,
heat gained in warming up = heat lost in cooling down
heat to melt ice + heat to raise water temp = heat lost by steam
For 1 mol = 18 g
(1 mol x 6020 J/mol) + (18 g (Tf - 0 C) (4.184 J/g.C) = (18 g) (100 - Tf C) (4.184 J/g.C)
6020 + 75.312Tf = 7531.2 - 75.312Tf
150.624Tf = 1511.2
Tf = 10.033 oC
So final temperature will be 10.03 oC
Determine energy lost by steam to cool to 0 oC
q = (mass) (dt) (Cp)
where,
mass = 1 mol x 0.018 kg/mol = 0.018 kg
dT = 100 - 0 = 100 oC
Cp = 4186 J/kg. C
Feed values,
q = (1 mol) (100) (4186)
q = 4.186 x 10^5 J
Calculate how much ice is melted by 4.186 x 10^5 J
we have, latent heat of fusion for water is 335,000 J/kg
= 4.186 x 10^5 J/335000 J/kg = 1.25 kg
So we will have 1.25 kg of ice melted at the end of process in the system
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