A calorimeter consists of an interior metal bomb (container) surrounded by 2.22 kg of water all contained within an insulated exterior wall. A sample of propane, C3H8, weighing 6.89g, was burned in the calorimeter with excess oxygen. As a result of the combustion, the interior temperature rose from 23.61*C to 26.88*C. It can be assumed that there is no heat loss to the exterior wall nor to the surroundings. The heat capacity of the metal bomb and contents is 337 cal/*C. The specific heat of water at 25*C is 0.998 cal/g*C. Calculate the following. A) The change in temperature [26.88*C-23.61*C=3.27*C] B)The heat absorbed by the water in calories. [Q=delta T(m)(Csp)… (3.27*C)[(2.22kg)(1000g/kg)](0.998cal/g*C)=7.24x10^3 cal] C)The heat absorbed by the metal bomb and contents in calories [(3.27*C)(337 cal/*C)=1.10x10^3 cal] D) The total heat absorbed by the calorimeter in calories [7.24x10^3 cal + 1.10x10^3 cal= 8.34x10^3 cal] E) The moles of propane burned [6.89 g C3H8 x 1mol/ (3*12.01 + 8*1.008)gC3H8 = .156 mol C3H8] F)The heat of combustion of propane in kcal/mol. G)The heat of combustion of propane in kJ/mol
Can you check parts a through e? I'm not sure if I did them correctly. I don't know how to do parts f and g. Thanks for the help!
I didn't work through the math but your solutions seem ok to me. For part f, heat of combustion from part D divided by 0.156 mol propane will give you cal/mol. Change that to kcal/mol. For part g, the easiest way is to convert cal to joules. There are 4.184 joules in 1 calorie, then redo the procedure in part f but use joules/mol and change to kJ/mol.
I'm not sure if I am suppose to add a negative one of the parts. Should both the heat absorbed by the water and heat absorbed by the metal bomb be positive? I thought a temperature increase means an exothermic reaction. It would be great if you could clarify this for me. Thanks for the help!
The combustion of propane IS an exothermic reaction. The heat produced by it heats the water and it heats the calorimeter; therefore, both of the heats absorbed (one by the water and the other by the calorimeter) must be added. The total heat absorbed by the two equals the total heat given off by the combustion. So the combustion is exothermic; the absorption by the water and container are endothermic. You may add a negative sign to part f and g as is done in the site below showing delta H/g for heats of combustion.
http://www.webmo.net/curriculum/heat_of_combustion/heat_of_combustion_key.html
Sure! Let's go through parts a to e to check your calculations:
A) The change in temperature: You correctly subtracted the initial temperature from the final temperature to get 3.27°C. Good job!
B) The heat absorbed by the water in calories: You used the formula Q = ΔT(m)(Csp), where ΔT is the change in temperature, m is the mass of the water, and Csp is the specific heat capacity of water. By plugging in the values correctly, you obtained 7.24x10^3 cal. This is correct.
C) The heat absorbed by the metal bomb and contents in calories: You multiplied the change in temperature by the heat capacity of the metal bomb and contents to get 1.10x10^3 cal. Well done!
D) The total heat absorbed by the calorimeter in calories: You correctly added the heat absorbed by the water and the heat absorbed by the metal bomb and contents to get 8.34x10^3 cal. Great job!
E) The moles of propane burned: You used the molar mass of propane correctly and divided the mass of propane by the molar mass to get 0.156 mol. This is correct.
Now let's move on to parts f and g.
F) The heat of combustion of propane in kcal/mol: To calculate the heat of combustion of propane, you need to divide the total heat absorbed by the moles of propane burned. So, to convert the total heat absorbed in calories to kilocalories, divide 8.34x10^3 cal by 1000. Then, divide the result by the number of moles of propane (0.156 mol). This will give you the heat of combustion in kcal/mol.
G) The heat of combustion of propane in kJ/mol: To convert the heat of combustion from kcal/mol to kJ/mol, multiply the value by 4.184. This conversion factor will give you the heat of combustion in kJ/mol.
I hope this helps! Let me know if you have any further questions.