# Isobaric question

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

Staff Emeritus
Gold Member
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.

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