Specific heat capacity of water

AI Thread Summary
An experiment was conducted to determine the specific heat capacity of water, yielding a value of 4.33 kJ/(kg.K), which is close to the theoretical value of 4.18 kJ/(kg.K). The discussion raised concerns about potential sources of error affecting the results, such as heat loss to the kettle and insulation issues. Participants emphasized the importance of detailing calculations and understanding the methodology used to arrive at the experimental value. The conversation highlighted the need for the experimenter to demonstrate effort in identifying errors and clarifying their calculations. Overall, the thread underscored the significance of accuracy and thoroughness in scientific experiments.
rthrbe
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I had an experiment to find the specific heat capacity of water. Materials are electric kettle, logger pro, 1 kg water, and set up the time to 240 seconds.

The experiment value i got is 4.33 kJ/(kg.K) which is closed to the waters specific heat capacity 4,18 kJ/(kg.K).

I wonder what could be the other factors that affect my experiment.
 
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rthrbe said:
I had an experiment to find the specific heat capacity of water. Materials are electric kettle, logger pro, 1 kg water, and set up the time to 240 seconds.

The experiment value i got is 4.33 kJ/(kg.K) which is closed to the waters specific heat capacity 4,18 kJ/(kg.K).

I wonder what could be the other factors that affect my experiment.
Welcome to the PF.

In the future, please fill out the Homework Help Template that you are provided when starting a schoolwork/homework thread here. It makes it much easier to give you help. :smile:

What do *you* think might be other factors and sources of error in your experiment?
 
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rthrbe said:
The experiment value i got is 4.33 kJ/(kg.K) which is closed to the waters specific heat capacity 4,18 kJ/(kg.K).

I wonder what could be the other factors that affect my experiment.

4.18 kJ/(kg⋅K) means it takes 4.18 kJ of heat energy to change the temperature of one kilogram of water by one kelvin. Can you think of any reason why it took you more than 4.18 kJ to do the same thing to each kilogram of water in your kettle?
 
Mister T said:
4.18 kJ/(kg⋅K) means it takes 4.18 kJ of heat energy to change the temperature of one kilogram of water by one kelvin. Can you think of any reason why it took you more than 4.18 kJ to do the same thing to each kilogram of water in your kettle?
Yeah i mean i could have some errors in my experiment, like the heat energy transferred (not sure w the term) not only to the water but also to the kettle. the electric kettle is made of plastic... is there any factors or errors?
 
rthrbe said:
... is there any factors or errors?

Do you means errors or factors that could account for difference between the value you measured and the value measured by the experts?

Well, since the values are different, then obviously the answer is yes, there are.

What do you think those factors or errors could be? Does the fact that your value is higher than theirs tell you anything?

Maybe if you show us the calculation you did we might be able to help you.
 
Mister T said:
Do you means errors or factors that could account for difference between the value you measured and the value measured by the experts?

Well, since the values are different, then obviously the answer is yes, there are.

What do you think those factors or errors could be? Does the fact that your value is higher than theirs tell you anything?
the specific heat capacity of water is 4, 18 kJ, that is the theoretical value, but on our experiment we got 4, 33 kJ, it is pretty close, and apparently there should be some kind of errors like loss of energy, something like that. i am not sure that's why i am asking for help
 
rthrbe said:
the specific heat capacity of water is 4, 18 kJ, that is the theoretical value, but on our experiment we got 4, 33 kJ, it is pretty close, and apparently there should be some kind of errors like loss of energy, something like that. i am not sure that's why i am asking for help

Since your value differs from theirs, you know you have some error. You say you're "not sure" but you are sure.

So, please see if you can respond in some other way.

In other words there are two possible questions here:

1. Does the fact that the two values are different mean there's some error?
2. What could be the sources of error?

The answer to the first question is yes. The answers to the second question lie in the details of how you calculated your value. That is why I asked you the questions I asked you in my previous replies. I can't answer you unless you answer me.
 
Mister T said:
Since your value differs from theirs, you know you have some error. You say you're "not sure" but you are sure.

So, please see if you can respond in some other way.

In other words there are two possible questions here:

1. Does the fact that the two values are different mean there's some error?
2. What could be the sources of error?

The answer to the first question is yes. The answers to the second question lie in the details of how you calculated your value. That is why I asked you the questions I asked you in my previous replies. I can't answer you unless you answer me.
i mean what could be the factors that affect my experiment, i mean why did i get 4, 33 instead of 4,18 there should be an explanation to that...
 
rthrbe said:
i mean what could be the factors that affect my experiment, i mean why did i get 4, 33 instead of 4,18 there should be an explanation to that...[/QUOT
Mister T said:
Since your value differs from theirs, you know you have some error. You say you're "not sure" but you are sure.

So, please see if you can respond in some other way.

In other words there are two possible questions here:

1. Does the fact that the two values are different mean there's some error?
2. What could be the sources of error?

The answer to the first question is yes. The answers to the second question lie in the details of how you calculated your value. That is why I asked you the questions I asked you in my previous replies. I can't answer you unless you answer me.
it can be somewhere in the electric kettle, something like that? insulation? idk
 
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rthrbe said:
it can be somewhere in the electric kettle, something like that? insulation? idk

What calculation resulted in 4.33? Can you show us? Can you explain what you did?
 
  • #11
rthrbe said:
i mean what could be the factors that affect my experiment, i mean why did i get 4, 33 instead of 4,18 there should be an explanation to that...
You need to start showing some effort in this thread, or it will be closed. We do not do your schoolwork for you -- that would be cheating.

We keep asking for you to put in some effort here. What do YOU think those sources of error could be?
 
  • #12
Mister T said:
What calculation resulted in 4.33? Can you show us? Can you explain what you did?
I used the logger pro, and the gradient was 0.3808. I've used the formula E= cm delta T ofcourse,
 
  • #13
berkeman said:
You need to start showing some effort in this thread, or it will be closed. We do not do your schoolwork for you -- that would be cheating.

We keep asking for you to put in some effort here. What do YOU think those sources of error could be?
Can you please read the threads too, i have already shown my solution but I haven't figure out yet what really might be the other cause why my value is little bit higher.
 
  • #14
You have not shown your solution, only the result of your solution. Telling us that you used the "gradient" on LoggerPro tells me that you made a graph in LoggerPro and used the slope. LoggerPro is software that let's you graph sensor readings. But what sensor readings did you graph? Temperature? Once you found the slope what did you do with that number to get your result?

Read back through this thread and look at every question I've asked you. Give us the answers.
 
  • #15
Mister T said:
You have not shown your solution, only the result of your solution. Telling us that you used the "gradient" on LoggerPro tells me that you made a graph in LoggerPro and used the slope. LoggerPro is software that let's you graph sensor readings. But what sensor readings did you graph? Temperature? Once you found the slope what did you do with that number to get your result?

Read back through this thread and look at every question I've asked you. Give us the answers.
it is the time vs temperature graph
 
  • #16
My patience has come to an end. If, the next time I read this thread I don't see at least an attempt to answer every question I've asked, I'll not respond. Otherwise all I can do is repeat my questions. Something I've already done ad nauseam.
 
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  • #17
Thread is closed due to the OP continuing to refuse to show some effort on her question.

@rthrbe -- Check your messages.
 

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