# HW8Sol - Chem Engr 178 HW 8 Solutions 1 A sample of...

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Chem Engr 178: HW 8 Solutions 1.) A sample of polyethylene that is cooled so rapidly that it doesn't crystallize has a density of 865 kg/m 3 while if the polyethylene were completely crystalline, it would have a density of 1000 kg/m 3 and a specific enthalpy of melting of 294 J/g. You are given a polyethylene (designated MDPE) with unknown crystallinity and measure its density to be 930 kg/m 3 . Find the weight fraction crystallinity, the volume fraction crystallinity and the enthalpy of melting you would measure using differential scanning calorimetry for 1 kg of MDPE. To find the volume fraction of crystalline material, 481 . 0 865 1000 865 930 3 3 = - - = - - = m kg m kg a c a tot c ρ φ The weight fraction of crystalline material is given by 517 . 0 481 . 0 930 1000 3 3 = × = = = m kg m kg V V x c tot c tot tot c c c The measured enthalpy of melting would be g J g J H x H pure m c m 152 294 517 . 0 , = × = Δ = Δ

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2.) Lets get some practice explaining concepts. Please make your answers concise: a) Why do polymers usually melt over a range of temperatures unlike low molecular weight materials which usually have a sharp melting point. Pages 285-288 provide a good explanation of the effects crystallinity on melting temperature. According to equation 4.34, 2 o o e m m m v T T T H l γ = - Δ The melting point of the polymer is highly depedenant on the lamellar thickness of the crystals. The crystallinity will never reach a 100%. Specifically, the smaller, localized regions of crystallinity will lead to a lower melting point while the larger crystalline regions will melt at a slightly higher temperature (relative to
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HW8Sol - Chem Engr 178 HW 8 Solutions 1 A sample of...

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