Thermodynamics - closed system

  • #1

Homework Statement



31.47mol of copper at 273 kelvin put inside an isolated cup along with 1 mol of water vapors at 373 Kelvin.
(pressure is constant at 1 atm).
ALL of the water condensed.
given parameters:
Cp(Cu(solid)) = 24.44 J/mol
Cp(H20(gas) = 33.58 J/mol
Cp(H20(liquid) = 73.35 J/mol
dHcondensation of water = -40,700J/mol.

a) find the final temperature of the system
b) what is the amount of heat transferred from copper to the water
c) find entropy change for water, copper and the whole system.

Homework Equations





The Attempt at a Solution



here is what im thinking.
This is isolated system so no heat is coming out.
the heat given from the water = the heat given from the condensation which is -40700J plus the heat given from cooling the water which is Q = Mass(water)*Cp(liquid)*(T(final) - 373K) = 75.35J/K (Tf - 373) Tf = final temperature.
Now the heat given from the copper (should be negative since the copper receive the heat from the water) is Q=Mass(copper)Cp(copper)*(Tf - 273) .
now since its isolated, the heat from the copper should be equal to the heat from the water but when summing it all up, i get that the final temperature is less than the starting temperature of the copper which is not possible ofcours.

thank you so much in advance.
 

Answers and Replies

  • #2
vela
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I think there's a problem with the numbers in the problem. Suppose all the heat released during condensation goes into heating up the copper. What's the final temperature of the copper? Does that answer make sense?
 
  • #3
The final temp of the copper should be higher than 273 k.
something is wrong with my sollution since i get lower than 273.
question is, where did i go wrong
 
  • #4
vela
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Ignore my previous post. I made a dumb mistake in my calculations.
 
  • #5
vela
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You just need to check your sign conventions.
here is what im thinking.
This is isolated system so no heat is coming out.
the heat given from the water = the heat given from the condensation which is -40700J plus the heat given from cooling the water which is Q = Mass(water)*Cp(liquid)*(T(final) - 373K) = 75.35J/K (Tf - 373) Tf = final temperature.
If 273 K < Tf < 373 K, you'll get a negative number for Q.
Now the heat given from the copper (should be negative since the copper receive the heat from the water) is Q=Mass(copper)Cp(copper)*(Tf - 273).
If 273 K < Tf < 373 K, you'll get a positive number for Q.
now since its isolated, the heat from the copper should be equal to the heat from the water but when summing it all up, i get that the final temperature is less than the starting temperature of the copper which is not possible of course.
When you put everything together, what was the equation you ended up with?
 
  • #6
D H
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I think there's a problem with the numbers in the problem.
The units are wrong. Specific heat should be in units of energy/mass/temperature. The numbers are correct assuming SI units of J/mol/K.

Suppose all the heat released during condensation goes into heating up the copper. What's the final temperature of the copper? Does that answer make sense?
The temperature rise in the copper due to just condensing the water vapor at 373 K to liquid at 373 K is surprisingly large. This is exactly why steam is so dangerous.


but when summing it all up, i get that the final temperature is less than the starting temperature of the copper which is not possible ofcours.
You have a sign error somewhere. Please write down the mathematical equations you used rather than trying to explain it in English.
 
  • #7
when i say the heat from the copper should be equal to the heat from the water i mean that in absolute value it equal but ofcours they have different signs since what is giving heat and one is gainning heat.
Q for water should be negative since they gave away heat.
Q for copper should be positive since copper gained heat
 
  • #8
when i say the heat from the copper should be equal to the heat from the water i mean that in absolute value it equal but ofcours they have different signs since what is giving heat and one is gainning heat.
Q for water should be negative since they gave away heat.
Q for copper should be positive since copper gained heat
 
  • #9
D H
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Right. So you should have ΔQwater+ΔQcopper=0.
 
  • #10
The units are wrong. Specific heat should be in units of energy/mass/temperature. The numbers are correct assuming SI units of J/mol/K.


The temperature rise in the copper due to just condensing the water vapor at 373 K to liquid at 373 K is surprisingly large. This is exactly why steam is so dangerous.



You have a sign error somewhere. Please write down the mathematical equations you used rather than trying to explain it in English.

It is not writted in the question but i assume Heat cap is per 1K.
here are math eqs.
For copper:
Q = M*cp*DeltaT = 31.47[mol]*24.44[J/mol*K] * (Tf-273) = 769.12*(Tf-273) (I)
For water
condensation
Q = -dH = -40700 J (II)
water cooling:
Q = m*cp*DeltaT = 75.35 (Tf-373) (III)

since isolated system : (I) = (II) + (III)
actually i get a pretty logic number for Tf if i put down plus sign in (II) instead of minus and maybe here is my mistake
 
  • #11
vela
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It is not writted in the question but i assume Heat cap is per 1K.
here are math eqs.
For copper:
Q = M*cp*DeltaT = 31.47[mol]*24.44[J/mol*K] * (Tf-273) = 769.12*(Tf-273) (I)
For water
condensation
Q = -dH = -40700 J (II)
water cooling:
Q = m*cp*DeltaT = 75.35 (Tf-373) (III)

since isolated system : (I) = (II) + (III)
actually i get a pretty logic number for Tf if i put down plus sign in (II) instead of minus and maybe here is my mistake
So are you saying your original equation was

769.12*(Tf-273) = -40700 + 75.35 (Tf-373)
 
  • #12
D H
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That is your mistake. It should be (I)+(II)+(III)=0.
 
  • #13
That is your mistake. It should be (I)+(II)+(III)=0.
if i do I + II + III = 0 i do get a logic answer, approx 330 K.
i see my mistake now

Thank you very much D H and vela!
 

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