Temperature of water electrolyzed at 5A for 1 min

You can't define the final temperature without knowing the initial temperature. So your approach is wrong, you should choose the initial temperature (e.g. 20°C) and then calculate the final temperature using ΔT.In summary, the given problem involves the electrolysis of 36mL of water at 15A/6V for 1 minute. Using the equations for Coulombs and Joules, the final temperature of the water can be calculated by subtracting the expanded gas from the initial mass and using the specific heat capacity formula. However, the final temperature cannot be accurately calculated without knowing the initial temperature. Therefore, using room temperature as the initial temperature, the final temperature can be estimated to be 55.969°C.
  • #1
HelloCthulhu
151
3

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=

shta.gif


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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  • #2
You can't calculate the final temperature not knowing the initial temperature.

Plus, what happens to water during electrolysis?
 
  • #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?
 
  • #4
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.
 
  • #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)

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
 
  • #6
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 °.
 
  • #7
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.
 
  • #8
HelloCthulhu said:
I'm trying to learn more about changes that occur

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

1. What is the effect of increasing the current on the temperature of water electrolyzed for 1 minute?

The temperature of water electrolyzed for 1 minute at 5A will increase compared to electrolyzing at a lower current. This is because a higher current increases the rate of electrolysis, leading to more heat being generated.

2. How does the temperature of water electrolyzed for 1 minute at 5A compare to other electrolysis conditions?

The temperature of water electrolyzed at 5A for 1 minute will vary depending on other factors such as the concentration of electrolyte, type of electrodes, and surrounding temperature. However, in general, higher currents and longer electrolysis times will result in higher temperatures.

3. Why does the temperature of water increase during electrolysis?

Electrolysis is a chemical reaction that requires energy, and this energy is converted into heat. Therefore, as the reaction proceeds, the temperature of the water will increase due to the heat generated by the process.

4. Can the temperature of water electrolyzed for 1 minute at 5A be controlled?

Yes, the temperature of the water can be controlled by adjusting the current, electrolysis time, and other factors such as adding a cooling system. However, it is important to note that higher currents and longer electrolysis times will inevitably result in higher temperatures.

5. Is there an optimal temperature for water electrolyzed at 5A for 1 minute?

The optimal temperature for water electrolyzed at 5A for 1 minute will depend on the specific purpose of the electrolysis. In some cases, a higher temperature may be desirable to increase the rate of electrolysis, while in others, a lower temperature may be preferred to minimize energy consumption and prevent the electrolyte from boiling.

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