1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Temperature of water electrolyzed at 5A for 1 min

  1. Jun 16, 2015 #1
    1. The problem statement, all variables and given/known data
    36mL of water undergoes electrolysis at 15A/6V. What is the temperature of the water after 1 min?

    2. Relevant equations

    1g = 1mL

    Coulombs = Amps x seconds
    Joules = Coulombs x Volt

    Specific Heat Capacity=


    3. The attempt at a solution

    15A x 60s = 900C

    900C x 6V = 5400

    5400J = (4.18 x 36g) x T

    5400/(4.18 x 36g) = 35.8 Celsius
  2. jcsd
  3. Jun 16, 2015 #2


    User Avatar

    Staff: Mentor

    You can't calculate the final temperature not knowing the initial temperature.

    Plus, what happens to water during electrolysis?
  4. Jun 16, 2015 #3
    I could use room temperature as the initial temperature:

    (T2-T1) = (35.8-20) = 15.8 °C

    During electrolysis,some of the water will turn into H2 and O2 gas. Should I subtract the expanded gas from the initial mass before solving for temperature?
  5. Jun 16, 2015 #4


    User Avatar

    Staff: Mentor

    Your room or my room? Question is impossible to answer, guessing information that was not given is not going to help. You can calculate ΔT but not the final temperature, period.

    Why do you subtract room temperature from the ΔT? That's not the correct approach. If the initial temperature was X and the temperature changed by ΔT, what is the final temperature?

    Yes. The difference is quite small, but high enough to change the ΔT.

    Actually what you can calculate has nothing to do with the real temperature change - some of the heat will run away with the produced gas, some will just get dispersed. All you can be sure is that the ΔT can't be larger than the one you have calculated.
  6. Jun 16, 2015 #5
    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)

    x +ΔT= T2?

    Not sure if this is the correct approach, but I'm trying to solve for final temperature with initial temperature 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
  7. Jun 16, 2015 #6


    User Avatar

    Staff: Mentor

    Looks reasonable, just watch sig figs in your final answer. And if you are so stubbornly decided to use 20°C as the initial temperature, you have to explicitly explain why you are doing this.

    When writing your message click on the ∑ above the edit field and you will be presented with a list of symbols, between them nicely looking °.
  8. Jun 16, 2015 #7
    I'm trying to learn more about changes that occur to a system during electrolysis. I'm just using room temperature as a benchmark for initial temperature.
  9. Jun 16, 2015 #8


    User Avatar

    Staff: Mentor

    Final temperature is not the change, the change is ΔT.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted