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Nut Calories – Pretzel Calories
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Mass = 4.12g
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Nut Calories – Pretzel Calories Mass = 4.12g Mass Al Cup= 27.65 g
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g Initial Temp =22 o C
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g Initial Temp 22 C Heat
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g Initial Temp 22 C Heat
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g Initial Temp 22 o C Heat
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al and water = 96.54g Initial Temp 22 o C Heat
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change) ( mass of water)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change) ( mass of water) (specific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released =( water temp change)( mass of water)(H 2 Ospecific heat capacity) + ( Al temp change) ( mass ofAl ) (specific heat capacity of Al)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released= (96.54g-27.65g)(48 C – 22 C ) ( 1 cal / g C) +
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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)
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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))
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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))
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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))
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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))
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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 x10 3 cal
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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 x10 3 cal
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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 x10 3 cal
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released per gram = 1.9 x10 3 cal / 4.12 g =
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released per gram = 1.9 x10 3 cal / 4.12 g =
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released per gram = 1.9 x10 3 cal =461.16 cal/g 4.12 g
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released per gram = 1.9 x10 3 cal =461.16 cal/g=4.6x10 2 cal/g 4.12 g
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released per gram = 1.9 x10 3 cal / 4.12 g = 461 cal / g=4.6x10 2 cal/g 461 cal/g ( 1 kcal / 1000 cal ) =.46 kcal
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C Heat Heat released per gram = 1.9 x10 3 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
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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
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Nut Calories – Pretzel Calories Mass = 4.12g Mass of Al =27.65g Mass of Al and water = 96.54g Initial Temp 22 o C Final Temp = 48 o C 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 %
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Results: The following data can be used to: Mass of cashew=1.34g Mass of aluminium cup = 19.83g Mass of aluminium cup and water=90.76g Initial temperature of Al cup and H 2 O=21.2 o C Final temperature of Al cup and H 2 O=68.8 o C Specific heat capacity of water = 1.0 cal / g o C Specific heat capacity of Al =.21 cal / g o C 1 Cal = 1 kcal = 1000 cal 1 cal = 4.18 J 1 kJ = 1000 J
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. 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= (m w t w * 1 cal / g o C ) + (m Al t Al *.21 cal / g o C ) Q = (90.76g – 19.83g) (68.8 o C – 21.2 o C) ( 1 cal/g o C) + ((19.83g) (68.8 o C – 21.2 o C) (.21 cal/g o C)) Q= 3574.48 cal 3570 cal 3.57x10 3 cal 2. Determine the number of calories per gram of food burned. cal = 3570 cal = 2664.179 cal / g 2.66x10 3 cal/g g (1.34g )
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3. Convert the number of calories per gram to Calories or kcal per gram. 2.66x10 3 cal/g ( 1 kcal) = 2.66 kcal / g (1000cal) 4. Convert the number of cal per gram to J per gram. 2.66x10 3 cal/g ( 4.18 J) = 11,118.8 J / g 1.11x10 4 J/g ( 1 cal) 5. Convert the number of cal per gram to KJ per gram 1.11x10 4 J/g ( 1 KJ ) = 11.1 KJ/g (1000 J)
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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 )
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