The heat of combustion of bituminous coal is 2.5 x 10^4 J/g. What quantity of coal is required to produce the energy required to convert 100. Pounds of ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius? Is the answer 5.36 about right. I'm not sure if I did it correctly
6 answers
Then why not post your work and let us check it?
How do I send a picture of my work here?
You type in the work like the rest of us.
Give me a second and I'll work the problem and see if 5.36 is about right.
Give me a second and I'll work the problem and see if 5.36 is about right.
100lbs x453.6g/lbs = 45360grams
45360 grams x 1mol/18.02 g= 2517.2 moles
2517.2mol x 6.01kj/mol = 15128.4kj
45360 grams x 4.18j/g C x 100C x 1/1000 = 18960.5
45360g x 1 mol/ 18.02g x 40.6 kj/ mol = 102198kj
All added equals 136287 kj
2.54E4 J /1g x #g = 136287 kj
#g= 5.36 kj
45360 grams x 1mol/18.02 g= 2517.2 moles
2517.2mol x 6.01kj/mol = 15128.4kj
45360 grams x 4.18j/g C x 100C x 1/1000 = 18960.5
45360g x 1 mol/ 18.02g x 40.6 kj/ mol = 102198kj
All added equals 136287 kj
2.54E4 J /1g x #g = 136287 kj
#g= 5.36 kj
I found your answer to my response last night and I expect you are ok. I used 2257 J for heat vaporization and 334 for heat fusion and 4.184 for specific heat H2O and came up with 5460 g,
Your 40.6 kJ/mol is about 2253 J/g and I used 2257.
I used 4.184 and you used 4.18.
Your 6.01 kJ/mol is 333.5 J/g and i used 334 for that. My answer came out 5460 g coal. Your procedure is good and my assumption is that the different numbers we used is responsible for the slight difference in g coal.
Your 40.6 kJ/mol is about 2253 J/g and I used 2257.
I used 4.184 and you used 4.18.
Your 6.01 kJ/mol is 333.5 J/g and i used 334 for that. My answer came out 5460 g coal. Your procedure is good and my assumption is that the different numbers we used is responsible for the slight difference in g coal.
Thank you very much!!!