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SCAN0032 - Chapter 12 Intermolecular Forces Liquids and...

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Unformatted text preview: Chapter 12 Intermolecular Forces, Liquids and Solids Refer to Chapter 5, 6, 9, 10 and 11 for review material as listed on page 425 (top right corner). Refer back to page 235. if we apply 22 J of heat to 4 g of water (Specific Heat = 4.184 J/gK) at 25°C, what will be the new T? if we apply 22 J of heat to 4 g of gold (Specific Heat = 0.129 J/gK) at 25°C, what will be the new T? Metals don’t have much of heat capacity, that is why water is used in radiators. That is great if we are not changing states, but water can be ice as well as a gas. Heat of fusion is the heat required to melt a solid to a liquid. Heat of vaporization is the heat required to boil the liquid into a gas. Since each of these occurs at their respective melting or boiling points, there is no change in T. Equation #1 qr = Hfm where m is the mass and Hf is the heat of fusion Equation #2 q.’ = va where m is the mass and H, is the heat of vaporization. For water these values are given on page 427. Hf = 333 Jlg = 6.02 kJ/mole H, = 2256 Jlg = 40.7 kJ/mole So, how much heat is required to move 22 g of ice at -22°C* to 122°C? Equation #3 QT = Q1 + Q2 4‘ Ola + Q4 + Cis = thiioe) + Cl? + thrwater) 1” Giv “i“ thisteam) Examine the Cooling Curve on page 429, Figure 12.3. Create a Heating Curve. Learn the names of the phase changes on page 428, Figure 12.2. ...
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