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5 Thermochemistry Solutions to Exercises (b) = - 3/2 O₂(g) = -393.5 kJ + 2(-241.82 -(-238.6 3/2(0) = -638.54 = -638.5 kJ (c) mol -638.54 kJ 1 mol 0.791 mL 1000 L mL = 1.58 10⁴ kJ/L produced 32.04 g (d) 1 mol CO₂ 44.0095 g = 0.06892 g emitted -638.54 kJ mol Foods and Fuels (Section 5.8) 5.81 (a) Fuel value is the amount of energy produced when 1 gram of a substance (fuel) is combusted. (b) The fuel value of fats is 9 kcal/g and of carbohydrates is 4 kcal/g. Therefore, 5 g of fat produce 45 kcal, while 9 g of carbohydrates produce 36 kcal; 5 g of fat are a greater energy source. (c) These products of metabolism are expelled as waste via the alimentary tract, primarily in urine and feces, and CO₂ (g) as gas. 5.82 (a) Fats are appropriate for fuel storage because they are insoluble in water (and body fluids) and have a high fuel value. (b) For convenience, assume 100 g of chips. 12 g protein 1g 17 protein kJ 1 Cal = 48.76 = 49 Cal 14 g fat 1g 38 fat Cal = 127.15 = 130 Cal 74 g carbohydrates 1g carbohydrates 17 kJ 4.184 1Cal = 300.67 = 301 Cal total Cal = (48.76 + 127.15 + 300.67) = 476.58 = 480 Cal % Cal from fat = 127.15 Cal fat = 26.68 = 27% 476.58 total Cal (Since the conversion from kJ to Cal was common to all three components, we would have determined the same percentage by using kJ.) (c) 25 g fat 38 g fat kJ protein g protein 17 kJ ; 56 protein 5.83 (a) Plan. Calculate the Cal (kcal) due to each nutritional component of the soup, then sum. Solve. 2.5 g fat fat = 95.0 or 0.95 10² kJ 17kJ 14 g carbohydrates = 238 or 2.4 10² kJ 1g carbohydrate 125