Temperature of water electrolyzed at 5A for 1 min

[HelloCthulhu]

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=

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

 

[Borek]

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

Plus, what happens to water during electrolysis?

 

[HelloCthulhu]

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?

 

[Borek]

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.

 

[HelloCthulhu]

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)

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

 

[Borek]

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 °.

 

[HelloCthulhu]

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.

 

[Borek]

I'm trying to learn more about changes that occur

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