Cubic and Linear Thermal Expansion.

[SherlockOhms]

Homework Statement

1. a 100m long copper wire of diameter 4mm is heater from 20C to 80C. What is the change in the length of the wire?
2. a 0.2m diameter aluminium sphere is cooled from 250C to 0C. What is the change in the volume of the sphere?
The coefficients of linear and cubic expansion were given for both materials.

Homework Equations

α(L) = dL/dT(1/L)
α(V) = dV/dT(1/V)

The Attempt at a Solution

For the first question is it not just a simple substitution exercise using the formula α(L) = dL/dT(1/L)? The fact that the diameter was provided is confusing me though.
For the second question isn't it just using the second equation α(V) = dV/dT(1/V), where V will be 4/3 * (pi) * r^3 and dT is -250?
Thanks, point out any incorrect observations!

 

[Chestermiller]

Homework Statement

1. a 100m long copper wire of diameter 4mm is heater from 20C to 80C. What is the change in the length of the wire?
2. a 0.2m diameter aluminium sphere is cooled from 250C to 0C. What is the change in the volume of the sphere?
The coefficients of linear and cubic expansion were given for both materials.

Homework Equations

α(L) = dL/dT(1/L)
α(V) = dV/dT(1/V)

The Attempt at a Solution

For the first question is it not just a simple substitution exercise using the formula α(L) = dL/dT(1/L)? The fact that the diameter was provided is confusing me though.
For the second question isn't it just using the second equation α(V) = dV/dT(1/V), where V will be 4/3 * (pi) * r^3 and dT is -250?
Thanks, point out any incorrect observations!

Your equations are the same as

\frac{d\ln{L}}{dT}=α_L
\frac{d\ln{V}}{dT}=α_V

You need to integrate these equations with respect to T to get the change in length or volume. Usually, when you do this, there will be a roundoff issue. You need to make use of the relation ln(1+x)≈x, or the relation exp(x)-1≈x to deal with this roundoff issue.

The diameter was deliberately included in the problem statement to confuse you and to test your understanding.

Your method for solving the second question is on target, and is appropriate for the first question also.

 

[SherlockOhms]

Brilliant. Thanks for the help!

 

[SherlockOhms]

Your equations are the same as

\frac{d\ln{L}}{dT}=α_L
\frac{d\ln{V}}{dT}=α_V

You need to integrate these equations with respect to T to get the change in length or volume. Usually, when you do this, there will be a roundoff issue. You need to make use of the relation ln(1+x)≈x, or the relation exp(x)-1≈x to deal with this roundoff issue.

The diameter was deliberately included in the problem statement to confuse you and to test your understanding.

Your method for solving the second question is on target, and is appropriate for the first question also.

Actually, would you mind explaining how that equation is integrated to give the equations which I gave above?

 

[Chestermiller]

Actually, would you mind explaining how that equation is integrated to give the equations which I gave above?

These are the equations you gave above, just re-expressed mathematically.

If you integrate these equations, you get:

ln(L/L_init)=α_L(T-T_init)

ln(V/V_init)=α_V(T-T_init)

If the change in L from its initial length is small, then you can write:

ln(L/L_init)=ln(1+(L-L_init)/L_init)≈(L-L_init)/L_init

The same goes for the change in V.

 

[SherlockOhms]

These are the equations you gave above, just re-expressed mathematically.

If you integrate these equations, you get:

ln(L/L_init)=α_L(T-T_init)

ln(V/V_init)=α_V(T-T_init)

If the change in L from its initial length is small, then you can write:

ln(L/L_init)=ln(1+(L-L_init)/L_init)≈(L-L_init)/L_init

The same goes for the change in V.

I see. Thanks for this!