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AdnamaLeigh
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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.
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.