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Daniel [21]
3 years ago
5

Suppose that all the dislocations in 4700 mm3 of crystal were somehow removed and linked end to end. Given 1 m = 0.0006214 mile,

how far (in miles) would this chain extend for dislocation densities of (a) 105 mm-2 (undeformed metal)? Enter your answer for part (a) in accordance to the question statement miles (b) 1010 mm-2 (cold-worked metal)?
Physics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

a. The required chain length = 292.1 miles

b. Length of the required chain = 29.2*10^6miles

Explanation:

Cold work is the work performed on a material at a temperature below the re-crystallization temperature of the material.

Dislocation density is the measure of the total number of dislocation of a crystalline material of unit volume.

Deformation is the process of change in an objects shape due to the application of force

a. In order to find the length of required chain for the given dislocation density

length of required chain = total length of dislocation line =  dislocation density * volume

where dislocation density = 10^5mm^{-2}

and volume is given as = 4700mm^3

=10^5*4700mm(\frac{1mi}{1.609*10^6mm} )=292.1miles

b. To find the length of required chain

length of required chain = dislocation density * volume

where dislocation density = 10^{10}mm^{-2}

we substitute in the equation above to get

=10^{10}mm^{-2}*4700mm^3\\=47*10^{12}mm*(\frac{1mile}{1.609*10^6mm} )=29.2*10^{6}miles

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Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

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J

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∘

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Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

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What if you wanted to increase the temperature of

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

4.18 J

+

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∘

C



4.18 J

=

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C



2

×

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To increase the temperature of

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n

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C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

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∘

C

increase in the temperature of

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

4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

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m

×

4.18 J

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m

grams of water by

n

∘

C

, you need to provide it with

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=

m

×

n

×

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n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

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m

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⋅

c

⋅

Δ

T

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T

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⋅

4.18

J

g

∘

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⋅

(

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−

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∘

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