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Temperature of gas during water electrolysis

  1. Jun 29, 2015 #1
    I know I can use the specific heat capacity formula to find the temperature of the water during electrolysis as long as I know the mass, specific heat constant of water and joules added beforehand:

    upload_2015-6-29_3-6-28.png

    But I'm not sure how I could find the temperature of the gas produced. If the pressure was given, I could use ideal gas law.

    8a1df8044b202c20aefce32e5ab98ecc.png


    But what if I only know the moles, volume of the container, and joules added? Any help would be greatly appreciated!
     
  2. jcsd
  3. Jun 29, 2015 #2

    Borek

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    That's not correct. Part of the energy was used for splitting water, not for heating it. (Which also means there is no straightforward answer to your further question).
     
  4. Jun 29, 2015 #3
    Can I subtract the gas produced from the mass and use that in the formula?
     
  5. Jun 29, 2015 #4

    Borek

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    No. Problem is not with the mass, but with the energy that was used for heating water. Actually we have ignored that part of the problem when you asked very similar question some time ago.

    Energy you put into the electrolysis is used for water decomposition and heating - heating of both water and escaping as. Assuming an isolated system calculating final temperature of the gas and water (they will be the same) is not that difficult. If the gas escapes this is by no means a trivial problem.
     
  6. Jun 29, 2015 #5
    So if it's a closed system, I can use the specific heat capacity formula for the water and the temperature will be the same for the gas?
     
  7. Jun 29, 2015 #6

    Borek

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    No, you will need to combine specific heat of water with mass water and specific heat of gas with gas mass. That will give you a heat capacity of the system, which can be used to calculate its final temperature, as ΔT will be identical.
     
  8. Jun 29, 2015 #7
    Let's use my last scenario (including a few adjustments):

    36mL of water at 20°C undergoes electrolysis inside a closed container at 15A/6V. What is the temperature of the water after 1 min? What is the temperature of the gas?

    Gas produced
    (15*60s*4g)/(F*4) = 0.00933g H2
    (15*60s*32g)/(F*4) = 0.07462g O2

    0.00933g + 0.07462g = 0.08395g

    Final mass
    36g-0.08395g = 35.916g

    Specific Heat Capacity
    5400J = (4.18 x 35.916g) x (T2-20°C)

    5400J = (4.18 x 36g) x ΔT
    5400/(4.18 x 35.916g) = (T2-20°C)
    35.969°C = T2-20°C
    55.969°C = T2

    Does this mean adding the specific heat of water (55.969°C) plus the mass (36mL)? And if I have to solve for the specific heat of produced gas, can I use the same value Q (5400J)? If not how can I calculate how much of the heat is now in the gas?
     
  9. Jun 29, 2015 #8

    Borek

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    What is the problem you are trying to solve? I have a feeling you are moving goalposts with each new thread (BTW: don't start new threads if it is the same problem all teh time).

    No, you can't add different things at random. Besides, 55.969°C is not a specific heat, but temperature, and 36 mL is not mass, but volume. You have to be precise about these things.

    Do you know what the heat capacity is? Do you know how it is related to the specfic heat capacity?
     
  10. Jun 29, 2015 #9
    Honestly I'm just trying to learn more about thermodynamics during electrolysis under standard and nonstandard conditions. I'll try to keep my questions to this thread from now on.

    Sorry, I often get a little confused. Heat capacity is the amount of heat required to raise the temperature of an object by 1°C
    and specific heat capacity is heat capacity per unit mass of a material. So when you say "combine specific heat of water with mass water" do you mean 4.18 plus 36g of water?
     
  11. Jun 29, 2015 #10

    Borek

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    By "combine" I didn't mean "add them". I meant "use them both to calculate heat capacity of water present".
     
  12. Jun 29, 2015 #11
    I'm sorry, but I either don't understand what you're asking for or haven't learned how to do that yet (or probably both). Are you asking me to solve for specific heat capacity for the heated water and solve specific heat capacity for the gas produced and then use both to solve for heat capacity of the system?

    If so, do I use the same number of joules in both specific heat equations? I'm very confused, but I'll make an attempt...

    Liquid:
    5400J = (4.18 x 35.916g) x ΔT
    5400/(4.18 x 35.916g) = (T2-20°C)
    35.969°C = T2-20°C
    55.969°C = T2

    Gas
    5400J = (2.080 x 0.08395g) x ΔT
    5400J/(2.080 x 0.08395g)=(T2-20°C)
    30.925°C=(T2-20°C)
    50.925°C=T2

    (55.969°C + 50.925°C)=106.894°C
     
  13. Jun 30, 2015 #12

    Borek

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    This is so unclear I have no idea whether you got it right or wrong.

    You don't "solve for specific heat capacities". You use them to calculate the heat capacity of the system. In the equation you have listed:

    [tex]Q = c \times m \times \Delta T[/tex]

    [itex]c \times m[/itex] is the heat capacity of the system consisting of one component. In general, for any system

    [tex]Q = c_s \times \Delta T[/tex]

    where cs is the heat capacity of the system. If the system consist of one component only

    [tex]c_s = m \times c[/tex]

    but if there are many components, you simply sum their individual heat capacities (beware: not specific heat capacities). So you can write system heat capacity as

    [tex]c_s = \sum_i m_i \times c_i[/tex]

    where mi and ci are masses and specific heat capacities of individual components. In your case this sum will consist of two parts - one describing the water, one describing the gas.
     
  14. Jul 1, 2015 #13
    Thank you so much for that clarification! I hope I understand how to do this now.

    Liquid 36mL x 4.18=150.48

    Gas 0.08395 x 2.080=0.17462

    150.48 + 0.17462=150.654

    [c][/s]=150.654

    5400J=150.654 x ΔT
    5400J/150.654=(T2-20)
    35.84+20=T2
    55.84=T2
     
  15. Jul 1, 2015 #14

    Borek

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    Don't omit units.

    While this gives a correct value, technically it is not correct.

    What are units of your result?
     
  16. Jul 1, 2015 #15
    Apologies. Thanks again for helping me out with this!

    Liquid 36mL x 4.18=150.48J/K
    Gas 0.08395 x 2.080=0.17462J/K
    150.48 + 0.17462=150.654J/K

    Cs=150.654J/K

    5400J=150.654 x ΔT
    5400J/150.654=(T2-20°C)
    35.84°C+20°C=T2
    55.84°C=T2
     
    Last edited: Jul 1, 2015
  17. Jul 1, 2015 #16

    Borek

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    You are still ignoring units - please write them everywhere.
     
  18. Jul 1, 2015 #17
    Liquid 36g x 4.18J⋅g−1⋅K−1=150.48J/K
    Gas 0.08395g x 2.080J⋅g−1⋅K−1=0.17462J/K
    150.48J/K + 0.17462J/K=150.654J/K

    Cs=150.654J/K

    5400J=150.654J/K x ΔT
    5400J/150.654J/K=(T2-20°C)
    35.84°C+20°C=T2
    55.84°C=T2
     
  19. Jul 1, 2015 #18

    Borek

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    So, you have silently corrected your initial mistake, didn't you?

    Please be careful about your notation, it is really not difficult to properly format your equations. We have tools for displaying heat capacity either as J⋅g−1⋅J⋅g−1⋅K−1 or [itex]J \times g^{-1} \times K^{-1}[/itex] or just [itex]\frac {J}{g \times K}[/itex]. Even J/(g⋅K) is better than things like g-1.
     
  20. Jul 1, 2015 #19
    Yes, I remembered what you said earlier about volume and mass. I'll be sure to use the thread tools on my next post. Thanks again Borek!
     
  21. Jul 1, 2015 #20

    Borek

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    There is nothing wrong with using the volume - just use density to convert to mass.

    [tex]36~mL \times 1 \frac g {mL} = 36~g[/tex]
     
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