FOOD ENGINEERING - QUESTION 23.1 : (a) According to United States Department of Agriculture (USDA) (http://ndb.nal.usda.gov/ndb/search/list, accessed 12 August 2016), 100 g of potatoes generate 77 kcal of energy. For raw tomatoes, 111 g have 18 kcal of energy. Question : How much energy will one gain if 150 g of heated potatoes are eaten with 200 g of raw tomatoes? (b) If 1 Calorie = 1 food Calorie = 1 kilocalorie and 1000 calories = 1 food Calorie, then how many Calories are there in 9600 calories? (c) According to a food package of potato chips, 210 Calories are produced per serving size of 34 g. In actual experiment of food calorimetry lab, 1.75 g of potato chips, when burnt, will produce 9.6 Calories. For each serving size of potato chip, find the difference of Calories between the actual experimental value and the value stated on the food package. (d) The specific heat of water is c = 1 cal / (g.K) where cal is calorie, g is gram and K is Kelvin. Then what is the temperature rise of water, in degree Celsius, when 150 g of water is heated by 9600 calories of burning food?
FOOD ENGINEERING - ANSWER 23.1 : (a) Energy gained = 150 g (77 kcal / 100 g) for potatoes + 200 g (18 kcal / 111 g) for tomatoes = 147.93 kcal. (b) 1 food Calorie = 1 Calorie = 1000 calories, then 9600 calories is 9600 calories (1 Calorie / 1000 calories) = 9.6 Calories. (c) Calories per serving size in actual experiment = 34 g (9.6 Calories / 1.75 g) = 186.51 Calories. Then difference of Calories = 210 - 186.51 = 23.49. (d) Temperature difference T in degree Celsius is same as Kelvin. Then c = 1 = 9600 calories / [ (150 g) T ] or T = 9600 / 150 = 64 degree Celsius. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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ENGINEERING DRAWING - EXAMPLE 18.2 : For a Vernier scale that consists of 0.11 cm per section, estimate the length on actual standard ruler of 0.1 cm per section for 2.39 m and 0.91 m.
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BIOCHEMICAL ENGINEERING INSTRUMENTATION - EXAMPLE 29.4 : The resolution of separation, Rs for chromatography is given by the formula Rs = (difference in retention time) / (average width at the base). In a chromatogram, 3 peaks a, b and c are found. Average widths W at the bases of the solutes are : Wa = 20 s, Wb = 40 s, Wc = 30 s. Resolutions of separation, Rs for solutes b and c in comparison to a are 2 and 4 respectively. The differences in retention times T for b and c in comparison to a are (Tb - Ta) and (Tc - Ta), Ta = Tc - Tb : (a) Form 2 equations involving Rs as a function of Wa, Wb, Wc, Ta, Tb and Tc. (b) Find the values of Ta, Tb and Tc.
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Question 41 - According to Raoult's law, the pressure when vapor is completely condensed, P = x(1) P(1) + x(2) P(2) ... + x(n) P(n) when x(1), x(2) ... x(n) are the mole fractions of component 1, 2 ... n and P(1), P(2) ... P(n) are the vapor pressures of component 1, 2 ... n. A vapor at 74 degree Celsius containing 70 mole % water and 30 mole % ethanol is to be completely condensed. At the temperature of 74 degree Celsius vapor pressure is 0.38 atm for water and 0.97 atm for ethanol. What is the minimum pressure the compressor must be operated?
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