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Melting/fusion

Melting/fusion. endo. endo. vaporization. endo. sublimination. freezing. exo. condensation. exo. exo. deposition. condensation. GAS. vaporization. LIQUID. FUSION. Freezing. Melting. SOLID.

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Melting/fusion

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  1. Melting/fusion endo endo vaporization endo sublimination freezing exo condensation exo exo deposition

  2. condensation GAS vaporization LIQUID FUSION Freezing Melting SOLID

  3. The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. Heat capacity = the amount of energy required to raise the temperature of an object (by one degree). • Molar heat capacity = heat capacity of 1 mol of a substance.

  4. Heat of fusion: The amount of energy released/required at the solid/liquid phase change. Heat of vaporization: The amount of energy released/required for liquid/ gas phase change.

  5. Heat is the total amount of energy possessed by the molecules in a piece of matter. This energy is both kinetic energy and potential energy. Temperature is proportional to the average kinetic energies of the molecules

  6. What happens when two objects of different temperatures come into contact? • moves spontaneously from matter with higher T to matter with lower T

  7. Which of the following has the greatest heat capacity? 100 g water or 1000 g water Which of the following has the greatest specific heat? 100 g water or 100 g copper

  8. How much heat is required to raise the temp of 205 g of water • from 15.2 C to 16.2 C ? What info do we have? Dt = 16.2 – 15.2 = 1 m = 205 g c= 4.184 J/g C What are we looking for ? q What is our equation ? q = mcDt q = (205 g) (4.184 J/g C) (1 C) 858 J

  9. Calculate the amount of heat released when 25 g of water • At 25 C is cooled to 0 C ? What info do we have? Dt = 25 – 0 = 25 C m = 25 g c= 4.184 J/g C What are we looking for ? q What is our equation ? q = mcDt q = (25 g) (4.18 J/g C) (25 C) 2615 J

  10. What mass of 67.5 C iron must be added to 235 g of 5.00 C water • To make the final temp of both come out to be 15 C ? What do we have? Iron c = .444 J/g C m = ? g Iron initial T = 67.5 C Iron final T = 15 C Dt =67.5 – 15 = 52.5 Water c = 4.184 J/g C Water m = 235 g Water initial T = 5 C Water final T = 15 C Dt= 15 – 5 = 10 C Heat lost = Heat gained q = q mcDt= mcDt (?g) (.444g) (52.5) = (235g)(4.184 J/g C)(10 C) mass = (235g)(4.184 J/g C)(10) (.444J/g C)(52.5 C) 421.8 g Fe

  11. A 195 g aluminum engine part at an initial temperature of 3.00 C • absorbs 40 KJ of heat. What is the final temperature of the part? What info do we have? T initial= 3.00 C m = 195 g c= .897 J/g C q = 40 KJ 40,000J What are we looking for ? T final What is our equation ? q = mcDt q= mc(Tfinal-Tinit) 40,000J = (195 g) (.897 J/g C) (Tfinal – 3.00 C) 40000J = Tfinal – 3.00 C (195g)(.897J/g C) 40000J + 3.00 C = Tfinal (195g)(.897J/g C)

  12. When 300 J of energy is lost from 125 g object, the temperature • decreases from 45 C to 40 C. What is the specific heat of this object? What info do we have? T initial= 45.00 C m = 125 g T final = 40C q = 300 J What are we looking for ? c specific heat What is our equation ? q = mcDt 300J = (125 g) (c) (5 C) 300J = c (125g)( 5 C) .48J/g C

  13. 6. The specific heat of lead (Pb) is 0.129 J/g C. Find the amount of heat released when 2.4g of lead are cooled from 37.2 C to 22.5 C? What info do we have? T initial= 37.2 C m = 2.4 g T final = 22.5C c = .129J/g C What are we looking for ? q What is our equation ? q = mcDt q = (2.4 g) (.129 J/g C) (22.5C – 37.2 C) 4.6 J

  14. 7. How many kJ of energy are needed to raise the temperature of 165 g water from 10.5 C to 47.32 C? What info do we have? T initial= 10.5 C m = 165 g T final = 47.32 C c = 4.18J/g C What are we looking for ? q in units of kJ What is our equation ? q = mcDt q = (165 g) (4.184 J/g C) (47.32 C – 10.5 C) = 25395 J 25395 J x 1 kJ = 25.4 KJ 1000 J

  15. 8. How much heat is absorbed by 15 g of ice being melted? H2O (s) H2O (l) DHf = 6.01 kJ/1 mol 15 g x 1 mol = .833 mol 18 g .833 mol x 6.01 kJ/mol = 5 kJ

  16. 9. How much heat is necessary to change 5.0 g of water at 100 C to to steam at 100 C? H2O (l) H2O (g) DHv = 2260 J/g 5g x 2260 J/g = 11300 J

  17. Calculate Molar Mass of Calcium Phosphate Ammonium Sulfate CaPO4 (40) + (31) + 4(16) = 135 g/mol (NH4)2SO4 (18)2 + 32+ 4(16) = 132 g/mol

  18. How many moles of Na2CO3 are thee in 10L of 2.0M solution 2.0 mol = X mol 1 liter 10 liter = 20 moles

  19. Find the molarity of a solution containing 59 g HCl In 500 ml of water. First convert grams to moles 59 g HCl x 1 mole HCL 36 g = 1.6 mol Molarity is moles per liter 1.6 mole x 1000 ml 500 ml 1 liter = 3.2M

  20. What volume (in ml) of 12.o M HCl is needed to contain 3.00 moles HCl 12.0 mol = 3.0 mol 1 liter X liter X = 3.0 12 X = .25 liter X = 250ml

  21. How many moles of gas would be present in a gas Trapped within a 100 ml vessel at 25 C at a pressure Of 2.50 atm n = PV RT PV = nRT 2.5atm (0.1L) (0.821)(298K)

  22. What volume will 1.27 moles of helium gas occupy at STP 1.27 mol x 22.4 L mol 28.5 L

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