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Good Morning 10/26/2014

Good Morning 10/26/2014. Today we will be going over the answers to the questions on Worksheet 11.2a. Tomorrow we will be doing Lab 18 and Friday Lab 19. So here we go. Worksheet 11.2a Specific Heat Capacity. 1. Complete the equation for finding energy transfer.  H = m •  T • C.

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Good Morning 10/26/2014

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  1. Good Morning 10/26/2014 • Today we will be going over the answers to the questions on Worksheet 11.2a. • Tomorrow we will be doing Lab 18 and Friday Lab 19. • So here we go.

  2. Worksheet 11.2a Specific Heat Capacity

  3. 1. Complete the equation for finding energy transfer. • H = m •T • C

  4. 2. What is the specific heat capacity for water? • C = 4.184

  5. 3. Calculate the energy required to raise the temperature of 750 g of water by 50.0 C. • H = • m = • T = • C = ? 750g 50.0 C 4.184 J/g•ºC  H = m • T • C 750g 4.184 J/g•ºC 50.0 C  H = 156,900 J = 156.9 kJ

  6. 4. Calculate the specific heat of an unknown metal when 250 g of the metal is cooled from 90.0C to 28.0C when added to 133 g of water whose temperature went form 22.0C to 26.0C. • This is just like the lab, where did the heat go? • Into the water so Hmetal = Hw

  7. 4. Calculate the specific heat of an unknown metal when 250 g of the metal is cooled from 90.0C to 28.0C when added to 133 g of water whose temperature went from 22.0C to 26.0C. • H = ? • m = 250 g • T = 62 ºC • C = ??? H = mw• Tw• Cw 133g 4.0 C 4.184 J/g•ºC =2225.9 J

  8. H m x T = C We need C all by itself H = m x C x T m x T m x T 2225.9 J = 0.1436 J/g· C 250.0 g 62C

  9. 5. Calculate the energy required to raise the temperature of 365.0 g of water by 25.0 C. ? • H = • m = • T = • C = 365g 25.0 C 4.184 J/g•ºC  H = m • T • C 365g 4.184 J/g•ºC 25.0 C  H = 38,179 J = 38.2 kJ

  10. H m x T = C 6. Calculate the specific heat (C) of an unknown metal when 135.0 g of the metal is cooled from 90.0 C to 25.0 C by losing 3,396 Joules of energy. 3,396 J • H = • m = • T = • C = H = m x C x T 135 g m x T m x T 65.0 C ?? 3396 J = 0.387 J/g· C 135.0 g 65C

  11. H m x C = ΔT 7. Assume 126 Joules of heat is added to 5.00 g of water originally at 23.0 C. What would be the final temperature of the water? 126 J • H = • m = • T = • C = H = m x C x T 5 g m x C m x C ? 4.184 J/g•ºC 126 J = 6C 5.0 g 4.184 J/g•ºC Final Temp= 29C

  12. You don’t have to do 8, it isn’t a great question. Sorry I didn’t delete it before handing it out.

  13. 9. If 75.0 g of butane is burned, how much thermal energy, in kilojoules, is produced. Butane’s molar heat of combustion = 2859 kJ/mol

  14. 10. How much thermal energy is produced when one gallon of octane is burned? Note: One gallon of octane has a mass of 2,660 g and the molar heat of combustion is 5450 kJ/mol

  15. Lab 18 PRE-LAB EXERCISE • In the space below, combine two of the equations algebraically to obtain the third equation. Indicate the number of each reaction on the shorter lines. • Hess’s Law says if you can combine 2 or more reactions to create a reaction you can add the H’s for those reaction s to equal the H of the final reaction. • It is a little tricky but if you look close Reaction (2) is a combination of Reactions (1) and (3) (1) NaOH(s)  Na+ + OH–H1 + + (3) Na+ + OH– + H+ + Cl– H2O(l) + Na+ + Cl–H3 = = (2) NaOH(s) + H+ + Cl– H2O(l) + Na+ + Cl–H2

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