Question

According to the internet (correctly described as a landfill, not a library) the heat capacity of polyethylene is about 1850 J/kg-K. According to the classic text by Billmeyer, the heat of fusion of polyethylene is about 66 cal/g. How many grams of ice at 0oC (assume it stays at this temperature after melting) will be required to cool and then crystallize two pounds of polyethylene? The initial temperature is 150oC. You should look up the crystallization temperature. From memory (even less like a library) the heat of fusion of ice is 80 cal/g.

Answer 1

Information used :

Crystallization temperature of polythene = 95^oC = 368K.....[Approx.; the exact temperature depends on the quality of the polythene crystallized]

Latent heat of fusion of polythene = 68.4 cal/g = 285.9 J/g ....[According to experiment conducted by Bernhard Wunderlich and C. M. Cormie of Department of Chemistry, Rensselaer Polytechnic Institute, Troy, New York .]

dH_fusion of ice = 334 J/g

Heat capacity of polythene = 1.850 J/g-K

2 pound = 907.18 g

Calculation:

Heat required to drawn out from polythene at 150^oC or 423K to crystalline polythene at 95^oC

= [907.18 g * 1.850 J/g-K * (423-368)K] + [907.18 g * 285.9 J/g]

= 351668 J

Mass of ice required

= 351668 J / (334 J/g)

= 1052.9 g