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# Lab Report Uncertainties help?

Hi! I need some help with calculating the uncertainty for my lab report.

If there is an uncertainty of ±1 second on a measurement of time in seconds ( e.g 120s ±1) and the time is converted to minutes, what does the uncertainty become? (2 minutes ±?).

Thanks

± 1/60 minutes?

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Essentially you shouldn't be trying that because you'd want to keep the number of decimal places constant for your measured value and uncertainty. I'd only convert to minutes in your case if it was an absolute requirement.

To understand how uncertainties work for this, try to think about what 120s ±1 actually means. 120 seconds plus or minus one second. When you convert to minutes you divide 120 by 60; the +/-1, which is still units of seconds, should be divided by 60 as well.

Geez, how do you guys use the uncertainty sign? All I can do is +/-. Do you just copy and paste?

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Essentially you shouldn't be trying that because you'd want to keep the number of decimal places constant for your measured value and uncertainty. I'd only convert to minutes in your case if it was an absolute requirement.

To understand how uncertainties work for this, try to think about what 120s ±1 actually means. 120 seconds plus or minus one second. When you convert to minutes you divide 120 by 60; the +/-1, which is still units of seconds, should be divided by 60 as well.

Geez, how do you guys use the uncertainty sign? All I can do is +/-. Do you just copy and paste?

My experiment is finding boiling points experimentally, I took a temperature reading every minute. Do you think I should put the time in minutes or seconds?

Yeah I copied pasted the sign

Thank you for explaining this for me

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Hi guys! I'm sorry I keep on asking, but my teacher is not much help

Am I doing this right?

∆H_combustion=Q_total/n_combusted

Q_total=Q_w+Q_c Q_w=mc∆t Q_c=mc∆t

n_combusted=(final mass of spirit burner-initial mass ofspirit burner )/(molar mass of alcohol)

Q_w=100.00×4.184×(60-19)=17154.4J

Q_c=85.05×0.385×(60-19)=1342.5J

Q_total=17154.4+1342.5=18496.9J=18.5kJ

Molar Mass of Butanol (C_4 H_9 OH)=(4×12.01)+(10×1.01)+ 16.00 = 74.14 g/mol

n_combusted=(188.98-188.35 )/74.14=0.0085 mol

∆H_combustion=18.5/0.0085=-(2176.5 kJ)∕mo l(±100)

Finding the uncertainty:

Fractoinal uncertainty of Q_w=0.5/100+1/41=0.029

Absolute uncertainty of Q_w=0.029×17154.4=497.5=±500

Fractoinal uncertainty of Q_c=0.01/85.05+1/41=0.025

Absolute uncertainty of Q_c=0.025×1342.5=33.6=±30

Uncertainty of Q_total=500+30=530=±500

Fractional Uncertainty of 〖∆H〗_combustion=500/(18.5×10^3 )+0.01/0.63=0.043

Absolute uncertainty of 〖∆H〗_combustion=0.043×2176.5=93.6=±100

Thank you!

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Hi guys! I'm sorry I keep on asking, but my teacher is not much help

Am I doing this right?

∆H_combustion=Q_total/n_combusted

Q_total=Q_w+Q_c Q_w=mc∆t Q_c=mc∆t

n_combusted=(final mass of spirit burner-initial mass ofspirit burner )/(molar mass of alcohol)

Q_w=100.00×4.184×(60-19)=17154.4J

Q_c=85.05×0.385×(60-19)=1342.5J

Q_total=17154.4+1342.5=18496.9J=18.5kJ

Molar Mass of Butanol (C_4 H_9 OH)=(4×12.01)+(10×1.01)+ 16.00 = 74.14 g/mol

n_combusted=(188.98-188.35 )/74.14=0.0085 mol

∆H_combustion=18.5/0.0085=-(2176.5 kJ)∕mo l(±100)

Finding the uncertainty:

Fractoinal uncertainty of Q_w=0.5/100+1/41=0.029

Absolute uncertainty of Q_w=0.029×17154.4=497.5=±500

Fractoinal uncertainty of Q_c=0.01/85.05+1/41=0.025

Absolute uncertainty of Q_c=0.025×1342.5=33.6=±30

Uncertainty of Q_total=500+30=530=±500

Fractional Uncertainty of 〖∆H〗_combustion=500/(18.5×10^3 )+0.01/0.63=0.043

Absolute uncertainty of 〖∆H〗_combustion=0.043×2176.5=93.6=±100

Thank you!

I couldn't understand much. First thing you'll be marked down, would be on the fact that you don't explain every single step that you're doing.

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I couldn't understand much. First thing you'll be marked down, would be on the fact that you don't explain every single step that you're doing.

I explained the calculation at the top of the page then did them, but it's not so clear when I copied and paste

I measured the delta h of enthalpy of butanol using calorimetry, but it's the uncertainties I'm not sure about.

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Uncertainty is equal to half of your smallest measurement, so for example if your smallest measurement of a measuring cylinder was 1ml, it would be +/- 0.5ml

Hope this helps, and good luck with your work!