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Aluminium bar

  • Thread starter speedy46
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Homework Statement



A 40mm diameter bar of aluminium is 2.5m long the bar is has its temperature raised from 20°c to 40°c. Determine the increase in length of the bar in mm. Take the Al = 24 × 10-6 °C-1
How much heat has been transferred into the bar if the SCH = 0.9kj/kg °C


Homework Equations





The Attempt at a Solution



Am stuck on where to start could any one help please
 

Answers and Replies

  • #2
Hootenanny
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Homework Statement



A 40mm diameter bar of aluminium is 2.5m long the bar is has its temperature raised from 20°c to 40°c. Determine the increase in length of the bar in mm. Take the Al = 24 × 10-6 °C-1
How much heat has been transferred into the bar if the SCH = 0.9kj/kg °C


Homework Equations





The Attempt at a Solution



Am stuck on where to start could any one help please
Saying that you are stuck on where to start isn't an attempt at a solution. You much have some ideas as to where to start. Have you read your class notes or course text?
 
  • #3
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increase in length = linear expansion coefficent × lenght of material (t2 - t1)

I do not understand where the t2 - t1 come from.

Is it the temperture increase - the temperture of the bar before it is rasied.
 
  • #4
tiny-tim
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increase in length = linear expansion coefficent × lenght of material (t2 - t1)

I do not understand where the t2 - t1 come from.

Is it the temperture increase - the temperture of the bar before it is rasied.
Hi speedy46! :smile:

Yes, it's the temperature increase - final temperature minus initial temperature.

Increase in length is more if the original length is more,

and it's more if the temperature increase is more.

So it's proportional to both of them (in other words, to their product). :smile:
 
  • #5
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would this also work for decreassing temperture on a material?
 
  • #6
tiny-tim
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would this also work for decreassing temperture on a material?
Yes … the effect is reversible.

If a particular temperature increase makes it longer, then the same decrease will return it to its original length. :smile:

And cooling it further will make it even shorter than its normal length.

Some armies used to keep iron cannonballs inside brass "barrels". If the temperature fell below freezing point overnight, the iron and the brass would both decrease in size, but the brass would decrease faster, so the cannonballs would get closer to the top, and sometimes one of them would even be forced out of the brass barrel! :biggrin:
 
  • #7
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please could you tell me if this right or not

increase in length = 24 × 10-6 × 40 mm (40 - 20) 24 ×10-6 × 40 × 20 = 0.0192
 
  • #8
tiny-tim
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please could you tell me if this right or not

increase in length = 24 × 10-6 × 40 mm (40 - 20) 24 ×10-6 × 40 × 20 = 0.0192
Looks good! :smile:
 
  • #9
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thank you now i have to work out how much heat has been transfered into the bar if the SCH= 0.9KJ/ KG degrees Celsius.

What does the SHC stand for please
 
  • #10
tiny-tim
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  • #11
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Thank you

To work out heat transfer

Q= (t1-t2)/(l÷Ka)

so 20 - 40 = 20

20 divide by 2.5m divide by ?

I do not know what the Ka stands for
 
  • #12
tiny-tim
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Hi speedy46! :smile:
Q= (t1-t2)/(l÷Ka)

so 20 - 40 = 20

20 divide by 2.5m divide by ?

I do not know what the Ka stands for
Sorry, I've no idea what Ka is … where did you get it from? :confused:

All you need can be found in the units of the SCH in the question:
SCH = 0.9kj/kg °C
/kg °C means per kilogram, and per degree.

So you find the number of kJ by multiplying 0.9 by the weight and also by the change in temperature … and nothing else! :smile:

btw, wikipedia gives the SCH as 0.9 J /g K … can you see why that's the same?
 
  • #13
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so the change in temperture is 20 degress 20-40=20

but the question does not give the weight or can i work this out.
 
  • #14
tiny-tim
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so the change in temperture is 20 degress 20-40=20

but the question does not give the weight or can i work this out.
hmm … that's very true! :rolleyes:

they only give you the volume, and you would need to know the density also. :cry:
 
  • #15
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density of aluminum is 2.7g/cm³

volume of the bar is πr^2*h π * 20 2 * 2500 = 3141592.654

I changed meters in to millmeters to keep them in the same units 1000*2.5= 2500mm
 
Last edited:
  • #16
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so now what do I need to do knowing the volume and density
 
  • #17
tiny-tim
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so now what do I need to do knowing the volume and density
mass = volume times density,

then SCH = 0.9kj per kilogram, and per degree. :smile:
 
  • #18
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2.7 3 times 33141592.654 = 61835968.21

is this correct
 
  • #19
tiny-tim
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2.7 3 times 33141592.654 = 61835968.21

is this correct
Sorry, you've lost me. :confused:

Write it out in full … and remember you need the mass in kg.

(and don't work to 10 significant figures! :rolleyes: )
 
  • #20
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volume of the bar is pie times 20 squared times 2500 = 3141592
The density of aluminum is 2.7g/cm3

Mass = volume times mass

mass = 3141592 times 2.7g/cm3

mass = 61835955.34

6.1kg

is this correct am unsure
 
  • #21
tiny-tim
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volume of the bar is pie times 20 squared times 2500 = 3141592
The density of aluminum is 2.7g/cm3

Mass = volume times mass

mass = 3141592 times 2.7g/cm3

mass = 61835955.34

6.1kg

is this correct am unsure
Use the same units that you'll need at the end:

volume of the bar in cm3 is π times 2 squared times 250 = 3142
The density of aluminum is 2.7g/cm3

mass = 3142cm3 times .0027kg/cm3 = … ? :smile:
 
  • #22
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Use the same units that you'll need at the end:

volume of the bar in cm3 is π times 2 squared times 250 = 3142
The density of aluminum is 2.7g/cm3

mass = 3142cm3 times .0027kg/cm3 = … ? :smile:

3142cm3 times .0027kg/cm3 = 8.4834 kg/cm3
 
  • #23
tiny-tim
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3142cm3 times .0027kg/cm3 = 8.4834 kg/cm3
erm … you mean 8.4834 kg. :rolleyes:

So the heat transferred in kJ is … ? :smile:

(btw, it's joules with a litlle "j", but J or kJ with a big "J" … like watts and W or kW, or newtons and N or kN :smile:)
 
  • #24
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so now i place that number in to the equation

0.9 times 8.4834 times 20 = 152.70 KJ
 
Last edited:
  • #25
tiny-tim
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:biggrin: Woohoo! :biggrin:

(but "kJ", not "KJ"! :smile: )
 

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