Temperature of water electrolyzed at 5A for 1 min

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Discussion Overview

The discussion revolves around the calculation of the final temperature of water after undergoing electrolysis at a specified current and voltage. Participants explore the implications of initial temperature, mass loss due to gas production, and the effects of heat dispersion during the process.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the final temperature of water after electrolysis, arriving at 35.8°C based on energy input.
  • Another participant points out that the initial temperature is necessary for calculating the final temperature, questioning the assumption of room temperature.
  • Some participants suggest using room temperature (20°C) as a benchmark for initial temperature, while others challenge this approach, emphasizing the need for clarity on the initial conditions.
  • Concerns are raised about the mass loss of water due to gas production (H2 and O2) during electrolysis, with calculations provided for the mass of gas produced.
  • There is a discussion about whether to subtract the mass of the gas from the initial mass of water before calculating the final temperature, with some arguing that it is a small but significant factor.
  • One participant notes that the heat produced may not fully translate to a temperature change due to heat loss to the environment and gas production.
  • Another participant emphasizes that the change in temperature (ΔT) is distinct from the final temperature, reiterating the importance of defining initial conditions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial temperature to use or the impact of gas production on the final temperature calculation. Multiple competing views remain regarding the assumptions and approaches to the problem.

Contextual Notes

Participants express uncertainty regarding the initial conditions and the effects of gas production on the calculations. There are unresolved questions about the assumptions made in the calculations and the implications of heat loss during electrolysis.

Who May Find This Useful

This discussion may be useful for students and individuals interested in electrolysis, thermodynamics, and the calculations involved in energy transfer during chemical processes.

HelloCthulhu
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Homework Statement


36mL of water undergoes electrolysis at 15A/6V. What is the temperature of the water after 1 min?

Homework Equations


[/B]
1g = 1mL

Coulombs = Amps x seconds
Joules = Coulombs x Volt

Specific Heat Capacity=

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

Plus, what happens to water during electrolysis?
 
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?
 
HelloCthulhu said:
I could use room temperature as the initial temperature:

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.

(T2-T1) = (35.8-20) = 15.8 °C

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?

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?

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.
 
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)

Borek said:
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?

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
 
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 °.
 
Borek said:
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.

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.
 
HelloCthulhu said:
I'm trying to learn more about changes that occur

Final temperature is not the change, the change is ΔT.
 

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