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Nut Calories – Pretzel Calories PowerPoint Presentation
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Nut Calories – Pretzel Calories

Nut Calories – Pretzel Calories

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Nut Calories – Pretzel Calories

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  1. Nut Calories – Pretzel Calories

  2. Nut Calories – Pretzel Calories Mass = 4.12g

  3. Nut Calories – Pretzel Calories Mass = 4.12g Mass Al Cup= 27.65 g

  4. Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g

  5. Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g

  6. Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g Initial Temp =22 oC

  7. Nut Calories – Pretzel Calories Mass = 4.12g Heat Mass of Al and water = 96.54g Initial Temp 22 C

  8. Nut Calories – Pretzel Calories Mass = 4.12g Heat Mass of Al and water = 96.54g Initial Temp 22 C

  9. Nut Calories – Pretzel Calories Mass of Al and water = 96.54g Initial Temp 22 oC Mass = 4.12g Heat

  10. Nut Calories – Pretzel Calories Mass of Al and water = 96.54g Initial Temp 22 oC Mass = 4.12g Heat

  11. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat

  12. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =

  13. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change)

  14. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change) ( mass of water)

  15. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)

  16. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)

  17. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)

  18. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)

  19. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released =( water temp change)( mass of water)(H2Ospecific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)

  20. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +

  21. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +

  22. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +

  23. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +

  24. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C)) + (( 27.65g ) (48 C – 22 C ) (.21 cak / g C)

  25. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C)) + (( 27.65g ) (48 C – 22 C ) (.21 cak / g C))

  26. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C)) + (( 27.65g ) (48 C – 22 C ) (.21 cak / g C))

  27. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C)) + (( 27.65g ) (48 C – 22 C ) (.21 cak / g C))

  28. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C)) + (( 27.65g ) (48 C – 22 C ) (.21 cal / g C))

  29. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) * ( 1 cal / g C)) + (( 27.65g ) * (48 C – 22 C ) *(.21 cal / g C)) = 1942.109 cal = 1.9 x103 cal

  30. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) * ( 1 cal / g C)) + (( 27.65g ) * (48 C – 22 C ) *(.21 cal / g C)) = 1942.109 cal = 1.9 x103 cal

  31. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released= ((96.54g-27.65g)(48 C – 22 C ) * ( 1 cal / g C)) + (( 27.65g ) * (48 C – 22 C ) *(.21 cal / g C)) = 1942.109 cal = 1.9 x103 cal

  32. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released per gram = 1.9 x103 cal / 4.12 g =

  33. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released per gram = 1.9 x103 cal / 4.12 g =

  34. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released per gram = 1.9 x103 cal =461.16 cal/g 4.12 g

  35. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released per gram = 1.9 x103 cal =461.16 cal/g=4.6x102cal/g 4.12 g

  36. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released per gram = 1.9 x103 cal / 4.12 g = 461 cal / g=4.6x102cal/g 461 cal/g ( 1 kcal / 1000 cal ) = .46 kcal

  37. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat Heat released per gram = 1.9 x103 cal / 4.12 g = 461 cal / g 461 cal/g ( 1 kcal / 1000 cal ) = .46 kcal Accepted Value from package = 110 Kcal / 28.3 g = 3.9Kal/g

  38. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat 461 cal/g ( 1 kcal / 1000 cal ) = .46 kcal Accepted Value from package = 110 Kcal / 28.3 g = 3.9Kal/g % difference = (experimental – accepted ) / accepted * 100

  39. Nut Calories – Pretzel Calories Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 oC Final Temp = 48oC Mass = 4.12g Heat 461 cal/g ( 1 kcal / 1000 cal ) = .46 kcal Accepted Value from package = 110 Kcal / 28.3 g = 3.9Kal/g % difference = (experimental – accepted ) / accepted * 100 % = (.46 Kcal/g -3.9 Kcal/g ) / 3.9 kcal/g * 100 = - 88 %

  40. Results: The following data can be used to:

  41. .Calculations / Conclusions • 1. The heat released by the nut was absorbed by the Water and the Al cup. • The amount of energy absorbed by a substance is dependent on its specific heat capacity, its mass , and its temperature change. • Q= (mwDtw * 1 cal / g oC) + (mAlDtAl * .21 cal / g oC) • Q = (90.76g – 19.83g) (68.8oC– 21.2oC) ( 1 cal/g oC) + • ((19.83g) (68.8oC– 21.2oC) ( .21 cal/g oC)) • Q= 3574.48 cal 3570 cal  3.57x103 cal • 2. Determine the number of calories per gram of food burned. • cal = 3570 cal = 2664.179 cal / g  2.66x103cal/g • g (1.34g )

  42. 3. Convert the number of calories per gram to Calories or kcal per gram. 2.66x103cal/g ( 1 kcal) = 2.66 kcal / g (1000cal) 4. Convert the number of cal per gram to J per gram. 2.66x103cal/g ( 4.18 J) = 11,118.8 J / g  1.11x104 J/g ( 1 cal) 5. Convert the number of cal per gram to KJ per gram 1.11x104 J/g ( 1 KJ ) = 11.1 KJ/g (1000 J)

  43. 6. According to the packaging the nut should release 250 kcal per 42 grams.(cashews) =5.95kcal/g a) What is your percent error in your experiment? % error = ( your value – accepted value ) * 100 = ( accepted value ) % error = (2.66 kcal/g – (5.95kcal/g))*100 = -55.3% error (5.95 kcal/g )