Linear thermal expansion and work done

[dogman1234]

Homework Statement

Hello all,
I need to find the change in internal energy (delta U) and work done (W) by an aluminum rod (from the expansion) as it is heated from 20C to 100C.

Initial Length = 2m
I found delta length = .00384m
diameter = .01m
radius = .005m
delta temp = 80C equivalent to 80K
coefficient of linear expansion = 24e^-5 K^-1
density of Al = 2700 kg/m^3

Homework Equations

delta L = (coefficient of linear expansion)(initial length)(delta temp)

Q = (mass)(specific heat)(delta temp)

Q = (delta U) + W

The Attempt at a Solution

I've found delta L to be .00345m
and Q to be 30,536.28J <----------Is this right? It seems high. (pi*.005^2)(2m)(2700kg/m^3)(900J/kgK)(80K) = Q

I'm not sure where to go from here. My professor gave us this equation:

(delta U) = mass(constant volume coefficient:Cv)(delta temp)

How can I use this? The process does not have a constant volume and it's not a gas.?

Also, what type of process is this? I feel that I do not have a very good understanding of the relationship of work and heat flow. If you could explain this to me I would be forever grateful.*Note: this problem is for learning purposes only. Not for credit. I will, however, be tested over this material next week.

 

[Rick88]

I've found delta L to be .00345m
and Q to be 30,536.28J <----------Is this right? It seems high. (pi*.005^2)(2m)(2700kg/m^3)(900J/kgK)(80K) = Q

That's not too high, it's fine.

The process is an isobaric expansion, that is an expansion taking place at constant pressure.



Regarding how to use the heat capacity, try maybe having a look at this:
http://en.wikipedia.org/wiki/Heat_capacity


R.

 

[dogman1234]

Thank you Rick for your help. I think I've finally figured it out.

 

[Rick88]

You're welcome :)