Calculate Work Done on 2.8kg Aluminum at Atmospheric Pressure

[AdnamaLeigh]

Given: the thermal coefficient of expansion = 2.4 x 10^-5 (degrees C ^-1)

A 2.8 kg block of aluminum is heated at atmospheric pressure so that its temperature increases from 20C to 43C. Find the work done on the aluminum. Answer in units of J.

I originally wanted to do W=P∆V. I could do ∆V=Voβ∆T, however, I do not know how to find original volume (I do know how to find β) since I was not given density.

I tried doing W=P∆V=Nk∆T but I don't even know if that rule is true. But this is what I did:

N = 2800 g Al x (1/26.98 g Al) x (6.022 x 10^23 molecules) = 6.25 x 10^25 molecules of Al

W = (6.25x10^25)(1.38x10^-23)(316-293) = 19837.5J

Work done on the aluminum = -19837.5J However, this is incorrect. Can someone please help? Thanks.

 

[chroot]

Do you know the specific heat of aluminum? It seems to me that the coefficient of expansion may well be a red herring, of no actual use to the problem.

- Warren

 

[FredGarvin]

I would try the W=PdV angle.

If you set an arbitrary initial volume of 1 in^3 and use the volumetric expansion of $$\frac{\Delta V}{V_o} = 3 \alpha \Delta T$$ to calculate your volumetric expansion.

You can then use standard atmospheric pressure in Lbf/in^2 to finish. It will be a small number, but that was the first idea that popped into my mind.

 

[Physics Monkey]

AdnamaLeigh,

You know the mass of your block, so in order to find the volume of the block you need only know the density at 293 K. The density can be found here: http://www.chemicalelements.com/elements/al.html