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July 30, 2021, 03:28:44 am

### AuthorTopic: Error, calculating error for specific heat capacity  (Read 528 times) Tweet Share

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#### C123456789L

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##### Error, calculating error for specific heat capacity
« on: June 12, 2021, 06:26:21 pm »
0
Heyo,

I'm being asked to show how I would calculate error for a specific heat capacity depth study using a calorimeter, in short the depth study is simply measuring the transfer of heat from a metal to water and working backwards to find the specific heat of the metal.

The formula is: Q = mcΔt

Q = energy transferred (J)
m = mass of the water (in kg) +/- 0.005 (measured using scales)
Δt = change in temperature of the water (in C) +/- 0.05 (measured using a calorimeter)
c = specific heat capacity for water (in J/kg/C) which is 4.1813 (im not sure what error I would include here as its an pre-determined value off the internet)

So now I think I can do this, again please let me know if this is wrong because I'm a year 11 student who has no idea what he's doing:

So I think, from what the internet has told me, I would simply calculate the percentage error of each variable (m, Δt and c) and simply add them up.

Worked example:

Δt is calculated using the following formula (T_final - T_initial = Δt), so im not sure if I have to add the percentage error of both readings or if I can leave it as ±0.05 either way I said:

Δt = 2.1±0.05 which is a ±2.5% error
m = 8.12±0.005 which is a ±0.0625% error
c = 4.1813±0.00005 which is 0.0011958% error

From here I plugged my values without their error into the formula:

Q = mcΔt
Q = 8.12 × 4.18 × 2.1
Q = 71.22736 J

I then added the error of all my components below:

0.0625 + 2.5 + 0.0011958 = 2.5636968

So my answer would be 71.22736±2.5636968%

But that just looks disgusting so I have like no idea at all

She mentioned something about propagation of error and linked me to this pdf:

https://www.animations.physics.unsw.edu.au/sf/toolkits/Errors_and_Error_Estimation.pdf

But like im gonna be honest I have no idea what I'm doing at all, the internet isn't being very helpful and I keep running into dead ends and my head hurts as well which sucks but yea

Please, I will be forever grateful if you assist me in my conquest.

Thanks, Casey

#### fun_jirachi

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##### Re: Error, calculating error for specific heat capacity
« Reply #1 on: June 12, 2021, 08:36:26 pm »
+4
A possible reason your answer looks 'disgusting' is because you're not using significant figures correctly. The number of significant figures you're using is (to put it mildly) slightly unnerving.

A few notes:
- There's no error for the specific heat capacity of water. It's a constant that has been verified through countless experiments; take it for granted
- You also cannot add percentage error. As an example, if I got 50% on two separate exams, I wouldn't have gotten 100% wrong and 100% right in total. It's also very dodgy to be doing this across variables (and across different units).

The error you're suggesting is part of random error (ie. all the stuff with the plus/minus signs). This is typically not factored into calculations but is reflected upon when you discuss your results. Usually the error that is relevant to you is calculated by $\frac{|\text{actual value - calculated value}| \times 100}{\text{actual value}}$. Here, your calculated value is just the one you've calculated above (without all the plus/minus stuff - ignore all of it). The actual value you can search up on the internet (from a reliable source, of course).

Hope this helps
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#### C123456789L

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##### Re: Error, calculating error for specific heat capacity
« Reply #2 on: June 16, 2021, 08:59:53 am »
+2
A possible reason your answer looks 'disgusting' is because you're not using significant figures correctly. The number of significant figures you're using is (to put it mildly) slightly unnerving.

A few notes:
- There's no error for the specific heat capacity of water. It's a constant that has been verified through countless experiments; take it for granted
- You also cannot add percentage error. As an example, if I got 50% on two separate exams, I wouldn't have gotten 100% wrong and 100% right in total. It's also very dodgy to be doing this across variables (and across different units).

The error you're suggesting is part of random error (ie. all the stuff with the plus/minus signs). This is typically not factored into calculations but is reflected upon when you discuss your results. Usually the error that is relevant to you is calculated by $\frac{|\text{actual value - calculated value}| \times 100}{\text{actual value}}$. Here, your calculated value is just the one you've calculated above (without all the plus/minus stuff - ignore all of it). The actual value you can search up on the internet (from a reliable source, of course).

Hope this helps

I really appreciate you taking the time to respond to my query,

I have handed the task in, and after researching error propagation I was successful in applying it in my task along with the advice you provided.

The problem I was having is that I didn't know the name of what I was applying, which was error propagation.

"In statistics, propagation of uncertainty is the effect of variables' uncertainties on the uncertainty of a function based on them. When the variables are the values of experimental measurements they have uncertainties due to measurement limitations which propagate due to the combination of variables in the function."

My teacher just wanted me to apply this concept to my depth study, which was relatively easy.

What you mentioned I think which is relative percentage error and is only applied at the end after you've finished calculating the result and have a final answer to compare.

I do admit if you didn't point it out I probably would have forgotten about it so I thank you.

Thanks, Casey