Frog-61 Project 001_025 (Private/Restricted Access)



Name ______________________________________________ Block ______________ Date ______________Chemistry 541 Worksheet Ch 10: Specific Heat, Heat of Fusion, Heat of Vaporization1)A 12 oz. can of soda weighs about 450 grams. How many joules are released when a can of soda is cooled from 25 degrees Celsius (room temperature) to 4 degrees Celsius (the temperature of a refrigerator). The heat capacity of liquid water is 4.18 J / gram . oC.q = (450 g)(21oC)(4.18J/goC) = -39,500 J2)How many joules are required to heat 250 grams of liquid water from 00 to 1000 C ?q = (250 g)(100oC)(4.18 J/goC)=104,500J3)How many joules are required to melt 100 grams of water? The heat of fusion of water is 6.01 kJ / mole.100 g H2O x (1mol/18g) = 5.5 molq=(5.5 mol)(6.01 KJ/mol)= 33.4 kJ4)How many joules are required to boil 150 grams of water? The heat of vaporization of water is 40.67 kJ / mole.150 g x (1 mol/18g) = 8.3 molq = (8.3 mol)(40.67 kJ/mol) = 339 kJ5)How many joules are required to heat 200 grams of water from 25 0C to 125 0C? The heat capacity of steam is 1.84 J / g . 0CThree equations :q = (200 g) (75oC)(4.18 J/goC) = 62,700 J = 62.7kJq = (11.1 mol)(40.67 kJ/mol) = 452 kJq = (200g)(25oC)(1.84 J/goC) = 9200J = 9.2 kJ62.7kJ + 452 kJ + 9.2 kJ = 522.9 kJ6)How many joules are given off when 120 grams of water are cooled from 25 0C to -250C? The heat capacity of ice is 2.09 J / g . 0C.3 equations:q = (120)(25oC)(4.18 J/goC) = 12500J = 12.5 kJq = (6.6 mol)(6 kJ/mol) = 40.1 kJq = (120g)(25oC)(2.09 J/goC)= 6300 J = 6.3 kJ12.5kJ + 40.1 kJ + 6.3 kJ = -58.9 kJ7)How many joules are required to heat 75 grams of water from -85 0C to 1850C? The heat capacity of steam is 1.84 J / g . 0C.5 equations: q=(75g)(85oC)(2.09J/goC)=13.3 kJq=(4.1mol)(6.01 kJ/mol) = 25 kJq=(75 g)(100oC)(4.18J/goC) = 31.3 kJq=(4.1 mol)(40.67 kJ.mol) = 169 kJq = (75 g)(85oC)(1.84 J/goC) = 11.7 kJ13.3kJ + 25 kJ + 31.1 kJ + 169 kJ + 11.7 kJ = 250 kJ8)How many joules are required to heat a frozen can of juice (360 grams) from -5 0 C (the temperature of an overcooled refrigerator) to 110 0C (the highest practical temperature within a microwave oven)?Use the heat capacity of steam, ice and water for the heat capacity of the OJ5 equations:q = (360g)(5oC)(2.09 J/goC) = 3.67 kJq=(20 mol)(6.01 kJ/mol)=120 kJq=(360J)(100oC)(4.18 J/goC) = 150 kJq=(20 mol)(40.67 kJ/mol)=813 kJq=(360 g)(10oC)(1.84 J/goC)= 6.6 kJ3.67kJ + 120 kJ + 150 kJ + 813 kJ + 6.6 kJ = 1093 kJ9)How many joules are released when 450 grams of water are cooled from 4 x 107 0C (the hottest temperature ever achieved by man) to 1 x 10-9 0C (the coldest temperature achieved by man). 5 equations:q=(450g)(3.9E7)(1.84 J/goC) = 3.2E7kJq=(25 mol)(40.67 kJ/mol) = 1017 kJq = (450g)(100oC)(4.18 J/goC) = 188 kJq=(25 mol)(6.01kJ/mol) = 150 kJq=(450g)(1E-9oC)(2.09 J/goC) = 9.4E-10 kJ3.2E7 kJ + 1017 kJ + 188 kJ + 150 kJ + 9.4E-10 kJ = -3.2E7 kJ10)How many joules are required to raise the temperature of 100 grams of water from -269 0C (the current temperature of space) to 1.6 x 1015 0C (the estimated temperature of space immediately after the big bang)?5 equations:q = (100 g)(269oC)(2.09 J/goC) = 56.2 kJ56.2 kJ + 33.3 kJ + 41.8 kJ + 226 kJ + 2.9 E 14 kJ =q=(5.5 mol)(6 kJ/mol) = 33.3 kJ2.9 E 14 kJq=(100 g)(100oC)(4.18) = 41.8 kJq = (5.5 mol)(40.67 kJ/mol) = 226 kJq = (100 g)(1.6E15oC)(1.84 J/goC)=2.9 E 14 kJ ................
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