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# Enthalpy question (specific heat capacity)? watch

1. Hey,

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
2. (Original post by Mooodle)
Hey,

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
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
3. (Original post by Mooodle)
Hey,

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
Where you given the mass/(density and volume) of liquid used inside the calorimeter?
4. (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
5. (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?
6. 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.
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.
8. 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.
9. (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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