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 :)

 

[rwisz]

*
Hello,

You are correct in your calculation of delta L and Q. The value for Q does seem high, but it could be due to the large temperature change and density of aluminum. It is always a good idea to double check your calculations and units to make sure they are correct.

To find the change in internal energy (delta U), you can use the equation provided by your professor: (delta U) = mass(constant volume coefficient:Cv)(delta temp). In this case, the mass is the density of aluminum multiplied by the volume of the rod (pi*r^2*L). The constant volume coefficient for aluminum is its specific heat at constant volume (Cv), which can be found in a table or calculated using thermodynamic data. This equation is applicable in this situation because the aluminum rod is not undergoing any significant volume changes, so the change in internal energy can be calculated using the specific heat at constant volume.

The type of process in this situation is a non-adiabatic, non-isobaric process. This means that heat is being transferred to the system (the aluminum rod) and the pressure is not constant. This type of process is also known as an isochoric process, where the volume remains constant.

The relationship between work and heat flow can be explained by the first law of thermodynamics: energy cannot be created or destroyed, only transferred. In this situation, the heat energy is being transferred to the aluminum rod, causing it to increase in temperature and expand in length. This expansion results in work being done by the rod, as it pushes against its surroundings. The work done by the rod is equal in magnitude to the heat energy transferred to it, as shown in the equation Q = (delta U) + W.

I hope this helps to clarify the relationship between work and heat flow. Good luck on your test!