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Chapter 10. Causes of Change. Practice problem #1. Determine the energy, in Joules, needed to increase the temperature of 20.00 g of water from 300.0 K to 365.0 K. Ch 10 Section #1. Practice problem #2.
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Chapter 10 Causes of Change
Practice problem #1 Determine the energy, in Joules, needed to increase the temperature of 20.00 g of water from 300.0 K to 365.0 K. Ch 10 Section #1
Practice problem #2 Determine the energy, in Joules, released as 1.00 Kilograms of water cools from 72.0 oC to 30.0 oC. Ch 10 Section #1
Practice problem #3 Determine the energy, in Joules, needed as 500.00 mL of water increases from 300.0 Kelvin to 370.0 Kelvin. (1.00 mL =1.00 gram for water) Ch 10 Section #1
Practice problem #4 Determine the energy, in Joules, released as 1.000 Liter of water Cools from 350.0 Kelvin to 310.0 Kelvin Ch 10 Section #1
Ch 10 Section #1 Practice problem #5 Determine the energy, in calories, needed as 2.00 x 104 milligrams of water cools from 85.0 oC to 5.00 oC.
Practice problem # 6 Calculate the energy required as 2.50 x 103 grams of ice melts Ch 10 Section # 1
Practice problem # 7 Calculate the energy required as 8.00 moles of ice melts Ch 10 Section # 1
Practice problem # 8 Calculate the energy released as 4008.0 grams of water freezes. Ch 10 Section # 1
Practice problem # 9 Calculate the energy required as 1.00 Kilomole of water freezes. Ch 10 Section # 1
Practice problem # 10 Calculate the energy required as 5.00 x 103 grams of water boils (vaporizes). Ch 10 Section # 1
Practice problem # 11 Calculate the energy required as 16.0 moles of water boils (vaporizes). Ch 10 Section # 1
Practice problem # 12 Calculate the energy released as 8.016 x 103 grams of steam condenses. Ch 10 Section # 1
Practice problem # 13 Calculate the energy released as 2.00 Kilomole of steam condenses. Ch 10 Section # 1
Ch 10 Section #1 Practice problem # 14 Calculate the energy required as 180.0 g of Ice at 0.0 oC is heated to Steam at 100.0 oC. 0.0 oC Ice 0.0 oC Water Step #1 0.0 oC Water 100.0 oC Water Step #2 100.0 oC Water 100.0 oC Steam Step #3 Total Energy Required = Step #4
Ch 10 Section #1 Practice problem # 15 Calculate the energy released as 5.00 moles of steam at 100.0 oC is cooled to 0.0 oC Ice. 100.0 oC Steam 100.0 oC Water Step #1 100.0 oC Water 0.0 oC Water Step #2 0.0 oC Water 0.0 oC Ice Step #3 Total Energy Released = Step #4
Ch 10 Section #1 Bonus Practice problem: Extra Credit Calculate the total energy required as 20.0 moles of Ice at 248.0 K is heated to Steam at 423.0 K. For full credit you must show all six equations and answers.
Bonus / Extra Credit x Ch 10 Section #1 Finish x Start
Molar heat capacity The energy as heat needed to increase the temperature of 1.00 mole of a substance by 1.00 Kelvin (K). Ch 10 Section #1 Note: Values of C can be found in your textbook on page 343
Practice problem # 16 Determine the energy as heat needed to increase the temperature of 20.0 moles of mercury by 15.0 K. The value of C for mercury is found on page 343. Ch 10 Section #1
Practice problem # 17 Determine the energy as heat removed as 50.00 moles of Iron cools from 1500.0 oC to 150.0 oC. The value of C for Iron is found on page 343. Ch 10 Section #1
Practice problem # 18 Determine the energy as heat needed as 3.60 x 102 grams of liquid water warms from 1.00 oC to 99.0 oC. The value of C for liquid water is found on page 343. Ch 10 Section #1
Practice problem # 19 Calculate the molar heat capacity (C) of carbon, given that 126.0 Joules were needed to heat 24.00 grams of carbon by 2.000 x 102 K Ch 10 Section #1
Ch 10 Section #1 Practice problem # 20 A 0.0700 mole of octane absorbs 3.50 x 103 J of energy. Calculate the temperature increase of octane if the molar heat capacity of octane is 254.0 J/K.mole
Ch 10 Section #1 Practice problem # 21 Explain in your own words how you would convert the Specific Heat (Cp) of a material to the Molar Heat Capacity (C) of the same material.