URGENT HELP ( heat energy) (1 Viewer)

[Twickel]

I need to do an experiment. Say I have 50ml of water and the change in temp is 15. I do the experiment twice once with a beaker once in a copper can.

With the copper can do I calculate it using 50x0.387x15 + 60x0.387x15 ( the second part is the mass of the can and 0.387 is the heat capacity of copper)
maybe the fisrt part of the equation should be 50x4.18x15 + 60x0.378x15?

and with the beaker i just do 50x4.18x15 correct?

THANKS!

 

[dood09]

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/thread

 

[minijumbuk]

The mass of the copper can is not included in the calculations.
If that was the case, what made you think that you didn't have to include it into the calculations when using a glass beaker?

You must remember than in the equation ΔH= -mCΔT, that:
m stands for the mass of the object that you are heating, and C stands for the specific heat capacity, which varies with each different material. There is nothing to do with the mass of the container. If you weigh the amount of rubbish in the rubbish bin, would you count the rubbish bin as part of the rubbish? (Don't say yes, I'll kill you)

So all you need to do, using the data you gave,

ΔH= 50 x 4.18 x 15 for beaker
ΔH= 50 x 0.387 x ΔT for copper

Since you used a different material, it sounds logical that ΔT for copper will be significantly higher than 15, because it transmitted energy much more efficiently, and thus there will be more heat transferred, which gives a higher final temperature value than the one obtained in water.

 

[Twickel]

my chem teacher is the once who said. So thats good.

so its just 50x4.18x15 ( for beaker)
50x0.87x?? ( for copper) but id like to heat them both to the same temperature because im testing which is more efficient ethanol or methanol.

 

[minijumbuk]

Yes, heat it to the same temperature then. You can still make the change in temperature = 15. But that would mean that you're burning less ethanol than methanol.

Note the difference between ΔH and ΔH_molar
ΔH is merely the heat change of the thing you heated. It doesn't tell you anything else.
ΔH_molar, however, tells you how much heat is absorbed/released PER MOLE OF FUEL.

So if you want to calculate ΔH_molar of methanol and ethanol:
Beaker: ΔH_molar(50 x 4.18 x 15)/(grams of methanol burnt/molar mass of methanol)

Copper container: ΔH_molar= 50 x 0.387 x 15 / (grams of methanol burnt/molar mass of methanol)

For ethanol:
Beaker: ΔH_molar(50 x 4.18 x 15)/(grams of ethanol burnt/molar mass of ethanol)

Copper container: ΔH_molar= 50 x 0.387 x 15 / (grams of ethanol burnt/molar mass of ethanol)

 

[Twickel]

Thanks, so I really want to know why my cheam teacher said it lie this

0.387x57x15 + 4.18x50x15 His reason was that we want to see how much energy has gone to the copper calorimeter and not directly to the water.

Also I have to talk about the accuracy of this experiment comparing my results ( heat of combustion and molar heat of combustion to the textbook results and other results)

Isnt that bad because maybe they used 1000ml of water and heated their water to 90 degrees see what I mean?

 

[Twickel]

so one mol of meth can produce up to 159 722 kj of energy? If 0.9g of fuel is used and the molar mass of meth is 46? Assuming we r using the beaker 15x4.18x50

 

[minijumbuk]

I don't understand your chemistry teacher.

 

[Twickel]

Me to Ill just do as you said and debate if they say its wrong because I was thinking of what you were saying when he said to add the mass of the can.

 

[Twickel]

Ok today I did the prac, I was heating 50ml of water from 18 degrees to 30 degrees using ethanol and methanol as fuels.

When the water was heated in the copper cup only 0.43g of methanol was used and only 0.28g of ethanol was used. However once the 50ml of water was heated in the beaker 1g of methanol was used and 0.45g of ethanol was used.

How can I explain these results?

Obviously the reason why a little amount of fuel was used when water was in copper was because it has a lower specific heat capacity hence the water heated up quicker, however its thermal conductivity is also high so more enery is used because we can assume that the copper cup temp increased by the same amount the water did.

When the water was heated in a beaker almost double the amount of each fuel was used, but its thermal conductivity is low so we can assum that all energy released by the fuels went to the water and not beaker.

Now how can I explain the differnce in fuel consumption?

From the results we can see that ethanol was much more efficient that methanol. Buring the experiment the amount of fuel burnt while heating 50ml of water in a beaker was almost double that of the fuel burnt while heating the same amount of water in copper. As we know copper is a metal, so it conducts heat better then glass.
Looking at the following calculations
∆h=mc∆t
Copper Calorimeter
(50x4.18x12)+61.41x0.387x12=2739
Glass Beaker
(50x4.18x12)+190x0.86x12=4423
In both those example it is assumed that both vessels change in temperature is the same as the temperature as the water contained in them. This can be said for copper because of its high thermal conductivity, however glass has a very low thermal conductivity hence we cannot say that the temperature of the beaker increased by 12 degrees, because the thermal conductivity of glass is so low we can say that the heat absorbed by the beaker is so little that it becomes negligible.
Having said that we now recalculate the our ∆h
Copper Calorimeter
(50x4.18x12)+61.41x0.387x12=2739
Glass Beaker
(50x4.18x12)= 2508
Specifically the copper vessel absorbed 285 of energy wheras the glass absorbed so little that it became negligible.
Now we can see that the glass beaker absorbed close to zero heat energy from both fuels , hence we can see why almost double the amount of each type of fuel was required to heat the same amount of water to the same temperature. Because the glass beaker absorbed less heat more heat enrgy was absorbed by other items e.g the metal clamps.


Thats my response to my question. Does that make sense is good?
__________________

 

[minijumbuk]

I have no idea what you just answered. It almost seems like a foreign language.

Sorry.

Would you mind rewording it? Take your time while typing out your response, and read it two or three times before submitting it. A few times, I can't even figure out what you're trying to say.