<p>Name: Block: Heat Problem Homework Table 1 Molar Heat Capacities of Elements and Compounds Element C(J/K.mol) Compound C(J/K.mol)</p><p>Aluminum, Al (s) 24.2 Aluminum chloride, AlCl3 (s) 92.0</p><p>Argon, Ar (g) 20.8 Barium chloride, BaCl2 (s) 75.1 Helium, He (g) 20.8 Cesium iodide, CsI (s) 51.8</p><p>Iron, Fe (s) 25.1 Octane, C8H18 (l) 254.0 Mercury, Hg (l) 27.8 Sodium chloride, NaCl (s) 50.5</p><p>Nitrogen, N2 (g) 29.1 Water, H2O (g) 36.8</p><p>Silver, Ag (s) 25.3 Water, H2O (l) 75.3</p><p>Tungsten, W (s) 24.2 Water, H2O (s) 37.4 Use Table 1 to answer the following questions 1-6. 1) Determine the change in temperature when 2.5 x 102 J of energy as heat is transferred to 0.20 mol of helium.</p><p>2) Use the molar heat capacity for aluminum chloride to calculate the amount of energy needed to raise the temperature of 260.5 g of aluminum chloride from 0C to 125C.</p><p>3) Calculate H when 1 mol of argon is cooled from 475 K to 233 K.</p><p>4) Calculate C for indium metal given that 1 mol In absorbs 53 J during the following process: In (s, 297.5 K)  In (s, 299.5 K)</p><p>5) Determine the specific heat of silver.</p><p>6) Determine the specific heat of sodium chloride. 7) How much heat needs to be removed to cause 112.0 g of barium chloride to experience a change of temperature from 15C to -30C if the specific heat of barium chloride is 0.36 J/K.g. </p><p>8) You need 70.2 J to raise the temperature of 34.0 g of ammonia, NH3 (g), from 23.0C to 24.0C. Calculate the specific heat of ammonia.</p><p>9) How many joules of heat are given off when 5.0 g of water cool from 75ºC to 25ºC, if the specific heat of water is 4.18 J/g.ºC?</p><p>10) How many joules does it take to melt 35 g of ice at 0ºC if the heat of fusion is 333 J/g?</p><p>11) How many calories are given off when 50 g of water at 0ºC freezes, if the heat of fusion is 79.72 cal/g?</p><p>12) a. How many calories are given off when 85 g of steam condense to liquid water? (Heat of vaporization = 539.4 cal/g)</p><p> b. Review question: How many Joules would that be?</p>