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I am currently doing an experiment where I am burning methanol to test for the heat of combustion. To test this, I am heating 100g of water and raising it by 20 degrees Celsius before extinguishing the flame.

Using $q=m c \Delta T$, I have calculated that I need 8.4 kJ of energy to heat 100g of water by 20 degrees Celsius. The initial amount of methanol was 3.96g, and the remaining after combustion was 3.59g. This means that 0.37g of methanol was used in the process. The heat of combustion was -726 kJ/mol.

I have forgotten to time my experiment to see how long it took for this to occur. I was wondering if there is a formula or relationship that I can use to calculate this so that I don't have to do the experiment all over again.

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  • $\begingroup$ The time taken to heat the water by 20 degrees, will depend on the method used, e.g. temperature of bunsen burner. $\endgroup$ – Tarius Jul 31 '15 at 12:02
  • $\begingroup$ This kind of test is normally done with precision equipment and a calorimeter. If you do this experiment in an open environment, a LOT of heat will dissipate into the environment, and your calculated answer will be WAY LOW. In addition, the data you are looking for is no doubt already published. $\endgroup$ – David White Jul 31 '15 at 18:00
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I am sorry to disappoint you, but there is no such formula that you can just apply. This is because it strongly depends on how and under what exact conditions and with wwhich tools you did the experiment.

Think of this: If every methanol molecule reacts (burns) at once all at the same time, then the exact same amount of energy is spent, but it went really fast. Depending on your container shape, the burning can have taken any amount of time and still the energy calculated is the same, even though the rate of the reaction is very different.

To find out more about reaction speeds, if all molecules were burned at the same time, you could ask that question in the chemistry SE forum, https://chemistry.stackexchange.com/.

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  • 1
    $\begingroup$ This is very true, now when I think about it :/ Damn, it was worth a try I guess... $\endgroup$ – user80922 Jul 31 '15 at 12:13

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