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Enthalpy question (specific heat capacity)?

Hey,

Could anyone please help me with this question:

A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715KJmol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

A step by step method and explanation would be much appreciated!

Cheers :smile:
Original post by Mooodle
Hey,

Could anyone please help me with this question:

A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715KJmol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

A step by step method and explanation would be much appreciated!

Cheers :smile:


1. work out the energy produced by burning 1 mol of methanol
2. what percentage is this of the enthalpy of combustion
3. apply the same percentage for the energy released by burning the propan-2-ol
Original post by Mooodle
Hey,

Could anyone please help me with this question:

A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715KJmol-1. The temperature of the calorimeter rose from 19.6°C to 52.4°C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol. 1.50 g of propan-2-ol CH3CH(OH)CH3 raised the temperature from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.

A step by step method and explanation would be much appreciated!

Cheers :smile:


Where you given the mass/(density and volume) of liquid used inside the calorimeter?
Reply 3
Avatar for Mooodle
Mooodle
OP
7
Original post by YouMadBro!
Where you given the mass/(density and volume) of liquid used inside the calorimeter?


No... the information in the question above was all the information provided :s-smilie:
Original post by charco
1. work out the energy produced by burning 1 mol of methanol
2. what percentage is this of the enthalpy of combustion
3. apply the same percentage for the energy released by burning the propan-2-ol


can you explain that method in simpler steps please?
I got that, from the experiments, 1 mol of methanol would lead to -11440kJ mol-1 . Is this the right sort of line?
I think I got it!
I wrote is down but thought it was dumb idea but now think it's the right answer

For methanol,
Enthalpy change = -(mass x heat capactiy x change in temperature) / mol
Putting the values in for methanol
-715 = -(mass x heat cap x 32.8) / (2/32)
-44.6875...= -32mc
1.363..=mc
Since you aren't given the mass of the liquid, I think you can take it to mean that it's included into the heat capacity as you can assume that the settings of the calorimeter does not change.

For propan-2-ol,
just put in the value you got for (mass x heat cap) to work out enthaply change.
Reply 6
Avatar for Mooodle
Mooodle
OP
7
Original post by YouMadBro!
I think I got it!
I wrote is down but thought it was dumb idea but now think it's the right answer

For methanol,
Enthalpy change = -(mass x heat capactiy x change in temperature) / mol
Putting the values in for methanol
-715 = -(mass x heat cap x 32.8) / (2/32)
-44.6875...= -32mc
1.363..=mc
Since you aren't given the mass of the liquid, I think you can take it to mean that it's included into the heat capacity as you can assume that the settings of the calorimeter does not change.

For propan-2-ol,
just put in the value you got for (mass x heat cap) to work out enthaply change.


Bless your soul sweet man.
May fortune smile upon you for the rest of your days.
I made an account on this website just to personally thank you for your genius. I have no idea how I didn't think to remove M from q=MS^T but thank you!
Original post by YouMadBro!
I think I got it!
I wrote is down but thought it was dumb idea but now think it's the right answer

For methanol,
Enthalpy change = -(mass x heat capactiy x change in temperature) / mol
Putting the values in for methanol
-715 = -(mass x heat cap x 32.8) / (2/32)
-44.6875...= -32mc
1.363..=mc
Since you aren't given the mass of the liquid, I think you can take it to mean that it's included into the heat capacity as you can assume that the settings of the calorimeter does not change.

For propan-2-ol,
just put in the value you got for (mass x heat cap) to work out enthaply change.
Original post by georgewood0
I made an account on this website just to personally thank you for your genius. I have no idea how I didn't think to remove M from q=MS^T but thank you!


Your welcome. I wish you the best of luck in the hell others call A-Level Chemistry. xD

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