Question

Paragraph Styles Font Section Please type your name here: Please be am work below. I would recommend wing your work on a piece of arch and then nping any work below. For dimensional analysis, please either insert an equation or we the notation below as an example for the common or 234 Tartan i Word select "Insert and then "Equation Data and Calculations: 1kg Record data for Steps 1-8 here. For steps 9-13, do the calculations below the chart first and then transfer data Food Cheetos Trial 1 Trial 2 Trial 3 234 9 10 2.202 lbs = 0.515 lbs 1 kg 234 gril ke 20 .202 lbs kg) 0.375 ha 1. Mass of Soda Can 41.28 41.28 41.2 2. Mass of Soda Can + Water 673 66,0 g 65.8 g Heat released by burning of food - heat absorbed by water in the can: (Eq 1) 3. Mass of Water (#2 - #1) 26.1 24.8 24.6 To calculate the amount of heat absorbed by the water when it is heated: (192) wale massime X wat X Temperaturewala 10.5 4. Mass of the Food + Paper Clip + Weigh Boat (before burning) 5. Mass of the Food + Paper Clip + Weigh Boat (after buming) the specific heat of water, which is calorie (g. C) 10,0 g 10.89 104 6. Mass of burned food (145) Calculations 1. Calculate the heat absorbed by water in calories (9) for each trial. Show work. Trial 1 0.5 0.3 0.4 7. Initial temperature of water 21.2°C 85.86 72.0'C 82.50 Trial 2 8. Final temperature of water 9. lleat absorbed by water in calories) 42 - Cale I below 10. Heat released by food (in calories) 41 11. Heat released by food (in Calories) Cale 2 below 12. Calories gram of food (16) Cale 2 below 13. Average Cal/g of food for Trial 3: 2. Show the calculation of the heat released by food in Caland Calg food for each trial. Trial : Trial 2: Trial 3:

need help with the calculations please.

Answer 1

Ans. * equation. qfood - qwater. ( Heal Released by = (heat absorbed by wates) Burning of food ) * To Calculate the amount of Heat Absorbed by the water khen it is heated. to colossos qwater = massewaters X (water) * Tempesture water) ean * Calculcetions. @ calculate the Heat Absorbed by water in calories fot each trical با ایران را را وا ا ر ا ا : له T * as we know, Heat absorbed by water, watermass X Cwares X AT given. (From table) = 26.1 x 1 x 64.7 m= 26.1 c=1 cal. 1688.67 cal OT=[Final-initial] = 85.8 - 21:1 = 64.70 * Trial 2! given m=24-8 c=1 OT = 72.0-21.20 = 50.8 qwater = mass x watex x AT = 24.8 X1 X 50.8 = 1259.84 cal

Trical 3 given m= 2460 24.6 c= 1 cal. OT= 82.5-21.1 = 61.4 & water = mass x cXAT. = 24.6 X 1 X 614 = 1510.44 cal ☺. Show The calculation of the heat Released by food in cal and cally food for each trial.

* trial 1: as we know, Heat Released by Burning of food = Heat Absorbed by water : q food - qwater for trial 1 q Food = 1688.67 col (from calculation@) (cally) food = 1688.67 -- -- from table. = 3377.34 callg. 0.5 * trial 2: - 1259.84 0.3 q food = 1259.84 cal. (cally) food = 1259.84 = 4199.46 cally tricel 3! a food = 1510.44 cal (cally) Food = 1510.44 & 0.4 = 3776.1 calls