Calculating Calorimeter Heat Capacity for Naphthalene Combustion

  • Thread starter Thread starter Taryn
  • Start date Start date
  • Tags Tags
    Heat
AI Thread Summary
To calculate the calorimeter heat capacity for naphthalene combustion, the user burned 136.8 mg of naphthalene, resulting in a temperature increase of 3.29 °C. They initially calculated the number of moles of naphthalene but struggled to determine the correct delta H value, questioning the accuracy of the heat of combustion they found online. The heat of combustion for naphthalene is noted as 5156 kJ/mol, which should be equated to the heat gained by the calorimeter. Despite attempts to apply this information, the user continued to encounter errors in their calculations. They plan to seek assistance from their lecturer for further clarification.
Taryn
Messages
63
Reaction score
0
You burn 136.8 mg of naphthalene in a calorimeter and the temperature of the calorimeter rises by 3.29 oC. Determine the calorimeter heat capacity in J K-1. The molar mass of naphthalene is 128.18 g mol-1

This is a question from a quiz I have to do.
this is wat I did.
Ccal=delta H/delta n=0.1368g/128.18g/mol
=1.06E-3moles

I don't know how to calc delta H from this info but I looked it up on the net and found 72.6+0.6kJ/mol.
Thats not right is it coz when I put this information in the formula
Ccal=((72.6+0.6)x1000j/molx1.06E-3)/(3.29+273.15)

I got it wrong... any help?
Does anyone know if my value for entropy of napthalene is right?
 
Physics news on Phys.org
Taryn said:
You burn 136.8 mg of naphthalene in a calorimeter and the temperature of the calorimeter rises by 3.29 oC. Determine the calorimeter heat capacity in J K-1. The molar mass of naphthalene is 128.18 g mol-1

This is a question from a quiz I have to do.
this is wat I did.
Ccal=delta H/delta n=0.1368g/128.18g/mol
=1.06E-3moles

I don't know how to calc delta H from this info but I looked it up on the net and found 72.6+0.6kJ/mol.
Thats not right is it coz when I put this information in the formula
Ccal=((72.6+0.6)x1000j/molx1.06E-3)/(3.29+273.15)

I got it wrong... any help?
Does anyone know if my value for entropy of napthalene is right?

heat of combustion of naphthelene=5156 kJ/mole. I think u may equate this to heat gained by calorimeter.
 
Thanks I will try that!
 
Yeah I just tried to use that and still got it wrong its so frustrating! I will go and see my lecturer now and hopefully he can help. Can u see anything wrong with my process?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top