Calculating Mass of Octane Needed to Heat Aluminum Block

[ghostanime2001]

Homework Statement

An aluminum engine block has a mass of 110 kg. If only 20% of the heat produced in the engine is available to heat the block, what mass of octane is required to raise the temperature of this block from 15 celcius to 85 degrees celcius? c(Al) = 0.9 kJ/(kg . degrees celcius)

Homework Equations

\DeltaH=mc\DeltaT

The Attempt at a Solution

\DeltaH=(110kg)(0.9)(85-15)=(110kg)(0.9)(70)=6930 kJ x 0.2 % = 1386 kJ

Im clueless as to find the mass of octane after this step :(

 

[Borek]

You will need specific enthalpy for the reaction of octane burning to solve the question.

 

[ghostanime2001]

The heat of reaction for the combustion of octane is -5517 kJ

 

[Joskoplas]

The heat of reaction for the combustion of octane is -5517 kJ

Nope. It's -5517 kJ/mol ... that "per mole" part is important. Presumably if you knew how much heat was produced, you could figure how how many moles of octane -- and if you knew that, from the molecular weight, how many grams of octane.

 

[ghostanime2001]

okay but none of them give me my answer: THe answer to this question is 0.72 kg

But anyways here's what I've tried to do

because only 20% of the heat produced is used to heat the block:

5517 kJ/mol x (1/1386 kJ) THe kJ part cancels out which gives me: 3.9805 mol and multiply that with 114 g/mol which gives me 453.77 g which is definently not the answer !

 

[Borek]

6930 kJ x 0.2 % = 1386 kJ

Apart from the fact that it is either 0.2 or 20%, but not 0.2%, are you sure have calculated here total amount of heat that have to be produced from the octane burning?

 

[ghostanime2001]

The entire Question goes like this:
Aviation gasoline is almost pure octane C_{8}H_{18}. Octane burns according to the equation:

\underbrace{C_{8}H_{18(l)}}_{a} + \underbrace{O_{2(g)}}_{b} \rightarrow \underbrace{CO_{2(g)}}_{c} + \underbrace{H_{2}O_{(l)}}_{d}


\DeltaH^{\circ}_{f} = -209 kJ for C_{8}H_{18},

\DeltaH^{\circ}_{f} = -286 kJ for H_{2}O,

\DeltaH^{\circ}_{f} = -394 kJ for CO_{2},

i) Balance the equation
C_{8}H_{18} + \frac{25}{2}O_{2} \rightarrow 8CO_{2} +9H_{2}O

ii)What is the heat of formation of reactant (b)
0 kJ because its an element

iii) Calculate the heat produced when 1.00 L of C_{8}H_{18} is burned. The density of C_{8}H_{18} is 703 g/L.

(-1) x 1. 8C_{(s)} + + 9H_{2(g)} \rightarrow C_{8}H_{18(l)} \DeltaH^{\circ}_{f} = -209 kJ

(9) x 2. H_{2(g)} + \frac{1}{2}O_{2} \rightarrow H_{2}O_{(l)} \DeltaH^{\circ}_{f} = -286 kJ

(8) x 3. C_{(l)} + O_{2(g)} \rightarrow CO_{2(g)} \DeltaH^{\circ}_{f} = -394 kJ


1. C_{8}H_{18(l)} \rightarrow 8C_{(s)} + 9H_{2(g)} \DeltaH^{\circ}_{f} = 209 kJ

2. 9H_{2(g)}) + \frac{9}{2}O_{2} \rightarrow 9H_{2}O_{(l)} \DeltaH^{\circ}_{f} = -2574 kJ

3. 8C_{(l)} + 8O_{2(g)} \rightarrow 8CO_{2(g)} \DeltaH^{\circ}_{f} = -3152 kJ
===========================================
C_{8}H_{18(l)} + \frac{25}{2}O_{2} \rightarrow 8CO_{2(g)} + 9H_{2}O_{(l)} \DeltaH^{\circ}_{c} = -5517 kJ

iv)Calculate the heat produced when 1.00 L of C_{8}H_{18(l)} is burned. The density of C_{8}H_{18(l)} is 703 g/L.
D = \frac{m}{V}

703g/L = \frac{m}{1 L}

m = 703 g

\frac{703g}{114g/mol} = 6.17 mol

\frac{5.52 x 10^{3} kJ}{mol} x 6.17 mol = 3.4x10^{4} kJ

v)An aluminum engine block has a mass of 110 kg. If only 20% of the heat produced in the engine is available to heat the block, what mass of octane is required to raise the temperature of this block from 15\circC to 85\circC degrees celcius? c_{Al} = 0.9 kJ/(kg \circC)

m=110 kg
c=0.9
\DeltaT=70 (85-15)
\DeltaH=(110)(70)(0.9)=6930 kJ (Total heat released in the aluminum block engine)

 

[ghostanime2001]

I don't understand what to do after part v)

 

[ghostanime2001]

Can somebody please help me out on this ? I've done all i can

 

[ghostanime2001]

An aluminum engine block has a mass of 110 kg. If only 20% of the heat produced in the engine is available to heat the block, what mass of octane(C8H18) is required to raise the temperature of the block from 15 degrees celcius to 85 degrees celcius? Specific heat capacity of aluminum = 0.9 kJ/(kg C)

The enthalpy of combustion of octane is -5517 kJ (the answer is 0.72 kg) But i don't understand how X_X

 

[ghostanime2001]

Does anyone even care about answering my question ?

 

[Ivan Seeking]

Everyone here donates their time. You will have to be patient.

 

[chemisttree]

Do it in pieces. How much energy will you need to raise the block by 1 degree C? Then multiply that by delta T, then account for the 20% efficiency.

You have the skills you need to answer this. Trust what you know and relax.

 

[Borek]

Do it in pieces. How much energy will you need to raise the block by 1 degree C? Then multiply that by delta T

That was already done

then account for the 20% efficiency.

And that's where the problems started...

ghostanime2001: as I already hinted at, amount of energy needed to heat the block if 20% is used for heating is NOT 6930 kJ x 0.2.

 

[ghostanime2001]

Then what do i do... ill say it again I've done ALL i can... i have no more ideas.

 

[chemisttree]

Borek, humor me.


Ghost, so tell me, how much energy do you need to raise the engine block by one degree C?

 

[ghostanime2001]

deltaH=(110)(70)(1) = 99 kJ

 

[chemisttree]

\DeltaH=mc\DeltaT

\DeltaT = 1 (if you want the energy per degree C)... you need 'c'

 

[ghostanime2001]

so i am right. u said the energy per degree I've given you the energy per degree man.. what else do u want yo WHy in the world would i need C man ? i don't need the heat capacity of octane dawg

 

[chemisttree]

Uhhhh, I guess. I just thought that you were trying to do too much at once and when you made a mistake it looked too complicated for you to figger out... dawg.

You should have written:

\DeltaH=(110 kg)(0.9 kJ/kg degrees C)(1 C)= 99 kJ (heat required to raise block by 1 C)

 

[ghostanime2001]

okay fine... now what

 

[ghostanime2001]

is this a calorimetry kinda question ? the aluminum engine block is the calorimeter and the octane in the substance acting like water ?

 

[chemisttree]

No, it isn't calorimetry.

Now you need to find the energy required to raise it 70 C. I think you know this already, so I'll tell you what comes after that. When you have that answer, you need to think of how much energy must be put into the engine at 20% efficiency to equal the energy required to raise the temperature by 70C. Like Borek said, that's where you made a minor error before...

Remember, you will need to add more energy than you need to raise the block by 70C if only 20% of it goes into heating the block.

You are going to kick yourself when you see how easy it is.

 

[ghostanime2001]

just tell me how to do it I am sick and ****ing tired of this question already

 

[chemisttree]

Ghost, you are obviously skilled, so you can do this. I'm trying to show you how to break down a problem into pieces when you are at your wits end, so bear with me.

You only made a minor mistake. You have to find it now.

 

[ghostanime2001]

find whaatttt

 

[ghostanime2001]

Im not lying i really do not understand how to do this question.

 

[ghostanime2001]

i really don't what more do u want me to say.

 

[ghostanime2001]

reword this question. Maybe its the language i don't understand.

 

[ghostanime2001]

im trying man I am ****ing trying !

 

[chemisttree]

Yes, you do. You made a super small error... you know your stuff but now you are pulling your hair out because of some trivial thing.

Your answer was 6930 kJ is required to raise the block by 70C. It isn't the heat released in the aluminum block engine. Its the energy absorbed by the block to raise the temperature by 70C. How much energy had to have been produced by the octane so that only 20% of it is equal to 6930 kJ?

Once you have that energy, you need to somehow convert it into mass of octane. Is there something you know about octane that tells you how much energy is produced per unit mass (or amount) that is burned?

 

[Bystander]

Yes, you do. You made a super small error... you know your (snip)

Amajor conceptual error --- reread the problem statement --- carefully. Ask yourself if there's a glass in your mother's kitchen into which you can pour an entire gallon of milk.

 

[ghostanime2001]

if what you say is true then i know 20% of energy is absorbed by the engine. THat 20% of energy released by octane is 6930. 100% of energy released by octane by burning would be 6930/0.2 = 34650 is this the total energy released by octane ? i assume so, and 20 percent of that is absorbed by octane which is 6930. Now how many moles is 34650 kJ ? uhhh i know 1 mole of combusion of octane is 5517 kJ/mol so 34650/5517 = 6.281 mol x 114g/mol molar mass of octane = 715.986 g

BUT THATS STILL WRONG... WHAT THE HECK... WHY DO SOME PROBLEMS ALWAYS SCREW ME UP AARRGGGGGGG

 

[Borek]

Even if it is wrong (numerically) you did this part correctly now.

Edit: it is not wrong, 715.986 rounds down to 720. Check signifficant digits throughout the problem.

 

[ghostanime2001]

................ I still don't know why i divided 6930 by 0.2

 

[ghostanime2001]

i did it only because i had no other choice to go. but i did it without understanding what I am doing.

 

[Borek]

I still don't know why i divided 6930 by 0.2

It follows from the % definition. There were X kJ, of which 20% were used to heat the block. Amount of heat used to heat the block was 6930 kJ. That means that

20% = 6930kJ / XkJ * 100%

Solve for X.

 

[chemisttree]

You found your original logical error. You should have divided instead of multiplying. That is minor from my POV. You could have caught it if you had used units in your calculations. Why did you divide 6930 (kJ absorbed by engine) by 20%? Look at the units if you did it the other way.

If the answer is wrong now it is either due to bad assumptions (like heat released per mole of octane burned) or arithmetic.

Try it on your own for awhile.

Edit. BTW, regarding whether to multiply or divide... think of it like this. I have a number (6930) that I am about to apply a factor to. I want the result to be bigger (arrived at logically). Do I multiply by a number less than one (20% or 0.2 in this case) or divide by a number less than one?

 

[ghostanime2001]

I don't understand

 

[ghostanime2001]

ive done all i tried but still don't understand.

 

[ghostanime2001]

All i know is that there's a freakin aluminum block engine. Heat is produced inside the engine from burning octane. 20% of that heat is used to raise the temperature of the aluminum engine block from 15 to 85 celcius. what mass of octane was used to raise the temperature of the engine block from 15 to 85 ?
q=6930 is the energy that raised the engine block from 15 to 85 so I DONT GET IT ..... T_TTTTTTTTTTTTTTTTTTTTTTT

 

[chemisttree]

Yes you do... Where did you get the value for the heat of combustion of octane?

DON'T GET SLOPPY WITH YOUR UNITS! Slow down.

 

[ghostanime2001]

The heat of combustion of octane is -5517 kJ from using Hess Law' and using heats of formation of CO2 and H2O in my first post.

 

[ghostanime2001]

is this question asking How much mass of octane provides 6390 KJ of energy that is 20% of the total energy produced from its total combusion inside the combustion chamber of the engine. So, how much mass of octane provides 6390 kJ of energy of which is used to heat the block from 15 to 85 celcius ?? what mass of octane delivers its 20% of energy from its total energy produced in the engine to the heating of the engine block ? In other words how many grams of octane provides 6390 kJ of energy that directly goes to heating up the aluminum engine block from 15 to 85

 

[chemisttree]

My bad, I thought that number was for energy per liter.

Carefully read post #34.

 

[ghostanime2001]

so what is it ? do i have the right idea now from post#44

 

[chemisttree]

You had it right in post #33. Borek caught it in #34.


720 g = 0.72 kg

 

[ghostanime2001]

okay so i got the right answer but that doesn't prove i understand the quetsion any better

 

[ghostanime2001]

it says what mass of octane can raise the temperature of the aluminum engine block from 15 to 85 not what the total mass of octane is present inside the combusion chamber of the engine block.

 

[Borek]

it says what mass of octane can raise the temperature of the aluminum engine block from 15 to 85 not what the total mass of octane is present inside the combusion chamber of the engine block.

Original question doesn't refer to the mass inside, but to the amount of octane that have to be burnt.

Think how car engine works - you put gasoline mixed with air in. It burns inside. Some of the energy is converted to work that you can get from the rotating crankshaft, some is lost in exhasut gases, some heats the engine block.

That's the same kind of situation - you are told how much temperature of the block rose, you are told what part of the heat was used to heat the block, you are asked how much gasoline (octane) was burnt. You are not told it was inside.