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6. Shown below is the heating curve for a substance. The five segments of the he

ID: 548425 • Letter: 6

Question

6. Shown below is the heating curve for a substance. The five segments of the heating curve are labeled A-E 130 100 T (°C) heat added (kJ) I. ndicate the phases of matter present when the substance is at each point (A-E) on the heating curve For this substance, which phase of matter has the largest heat capacity? How can you tell? In what state of matter would this substance exist at room temperature? How can you tell? IV. In terms of kinetic and/or potential energy, describe the energy changes that take place during segment C as heat is added to the substance In terms of kinetic and/or potential energy, describe the energy changes that take place during segment D as heat is added to the substance V. 7. Determine the amount of heat required to completely convert 46.0 grams of ethanol at 25°C to the vapor phase at 78°C. The information in the table below may be needed for your calculations. Please show all work. -114 C 78°C 5.02 kJ/mol 38.56 kJlmol 0.97 Jig-K 2.3 J/g-K 8. If an empty aluminum can weighs 13.49 g, how much energy does it take to completely melt 12 cans that are initially at 20.0°C? Aluminum completely melts at 660.0°C. Please show all work and use the information in the table below to help with your calculations s) = 24.2 J/mol°C -29.3 J/moloC AHfus = 10.71 k/mol | LAH,ap-294.0 kJ/mol

Explanation / Answer

7) heat required(q) = m*Cliquid*DT+n*DHvap

          m = mass of ethanol = 46 g

          Cliquid = specific heat of ethanol = 2.3 j/g.k

          DT = (78-25)

          n = no of mol of ethanol = 46/46 = 1 mol

         DHvap = 38.56 kj/mol

q = 46*2.3*(78-25)+(46/46)*38.56*10^3

    = 44.167 kj


8) heat required(q) = n*C(s)*DT+n*DHfus

    m = mass of aluminium = 12*13.49 = 161.88 g

    n = no of mol of aluminium = m/M = 161.88/27 = 6 mol

    C(s) = molar heat capacity of aluminium = 24.2j/mol.k

          DT = (660-20)

     DHfus = 10.71 kj/mol

q = 6*24.2* (660-20)+6*10.71*10^3

    = 157.19 Kj

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