Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Blair12 on December 18, 2016, 06:02:55 PM
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Calculate the mass of arachidonic acid needed to warm a 500 kg bear from 5°C to 25°C. (Assume that the average heat capacity of bear flesh is 4.18 J g-1 K-1)
What I've got so far is:
q=mcdelta-T
500kgx4.18x20K= 41'800J= 41.8KJ.
I'm stuck on what to do after this
Combustion equation is: C20H32O2 + 27O2--> 20CO2 + 16H2O
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You need enthalpy of combustion of the acid. Was it given?
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500kgx4.18x20K= 41'800J= 41.8KJ.
Check this calculations
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You need enthalpy of combustion of the acid. Was it given?
No, that's where I am confused. The acid is arachidonic acid
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500kgx4.18x20K= 41'800J= 41.8KJ.
Check this calculations
?
500x4.18=2090
2090x20=41800
41800/1000=41.8
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500kgx4.18x20K= 41'800J= 41.8KJ.
Check this calculations
Do I need to convert the Kg to grams before the calculation because that is what heat capacity is in (g^-1K^-1)?
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Check by the dimensional analysis.
Find enthalpy of combustion by searching internet.
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You need enthalpy of combustion of the acid. Was it given?
No, that's where I am confused. The acid is arachidonic acid
Is there a table of combustion enthalpies somewhere in the book?
In what context (what are you studying right now) was the question asked?
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This problem can be found in Olmsted Chemistry (Ch6 Energy and its Conservation) under group study problems in 2005 year or earlier.
https://books.google.pl/books?id=1vnk6J8knKkC&pg=PA383&lpg=PA383&dq=arachidonic+acid+enthalpy+of+combustion&source=bl&ots=WH4VqhhX39&sig=ILRc8bpF9mb-wX4ivqIQsmbueLM&hl=pl&sa=X&sqi=2&ved=0ahUKEwjM4LjS-__QAhXCBSwKHaZIBFIQ6AEIPDAE#v=onepage&q=arachidonic%20acid%20enthalpy%20of%20combustion&f=false
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Check by the dimensional analysis.
Find enthalpy of combustion by searching internet.
I found it on the internet at ΔH= -252.06. Once I found it I did: nacid= 41800KJ/252.06g KJ mol^-1= 165.83 mol
macid= 165.83mol x MM(304.47g mol^-1)= 50492g (1kg/1000g)= 50.49 Kg.
Is this correct? I converted the bear weight from Kg to g in the q=mcΔt calculation and am not sure if that is correct.
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ΔH= -252.06. [units missing]
Are you sure that this value concerns combustion?
Olmsted gives enthalpy of formation (see: google books) and student should calculate enthalpy of combustion using thermochemical data (for CO2 and H2O) from appendix. His answer is close to 1kg (this make sense since bears hibernate a few months long with continuous loss of heat and weight).
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ΔH= -252.06. [units missing]
Are you sure that this value concerns combustion?
Olmsted gives enthalpy of formation (see: google books) and student should calculate enthalpy of combustion using thermochemical data (for CO2 and H2O) from appendix. His answer is close to 1kg (this make sense since bears hibernate a few months long with continuous loss of heat and weight).
I Just did a quick internet search for the enthalpy of formation so i'm not 100% sure it was correct. What do I do after the q=mcΔT equation? I'm just confused on where to go after here
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Check google book link - ΔH0f = -636 kJ/mol
After calculation of enthalpy of combustion you should calculate how many moles fits to the heat needed (41800 kJ), then calculate mass of arachidonic acid. That's all.
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Books give tables of the heat of formation rather than combustion because from a "limited" (=huge but always insufficient) dataset you can deduce the heat of many more reactions, including combustions.
But you are supposed to make this deduction. That is, find the heat of formation of all reactants and products, and combine them to get the heat of reaction - here a combustion.
In this operation, especially for combustions, check carefully the aggregation state of the reactants and products, notably water, which can be liquid or gaseous. In a boiler it can be both, and this makes a difference in the harvested heat and the fuel efficiency. In a bear it must be both at the same time, since some water is excreted and some is perspired including in breath, so check for additional information, or maybe decide that fur animals perspire little.
Irrelevant in this present question, but important elsewhere: keep in mind that the heat of formation refers to the elements in their standard state, not to separated atoms.