Question

Some scientists believe that coronavirus (Covid-19) can not tolerate high temperatures. They say that keeping the food in hot water (above 70 °C) for at least 5 minutes will make the food safer compared to the uncooked food. a) Calculate the heat, in kilojoules, needed to prepare 10 L of hot liquid water at 80 °C if the initial temperature of water is 30 °C? b) A minimum of how many grams of isooctane (C3H18) must be burned to provide the necessary amount of heat in part (a) 2 C3H18 (1) + 25 O2(g) → 16 CO2(g) + 18 H20 (1) AH° = -1.09x104 kJ (d (H20 (1) =1.0 g/mL, Molar mass of C3H18 = 114.2 g/mol) Answers: a) kJ b) (write your answer with minimum of 3 significant figures) Constants: s (H20 (1) = 4.184 J/g. °C R = 8.314 J/K.mol or R = 0.08206 L.atm/K.mol 1 atm = 101.325 kPa h = 6.63x10-34 Js C= 3.0x108 m/s

Answer 1

of water ginen volume TOL= 10,000 ml man of water density xvolume 1.0 X10,000/= 10,00 2,000 g heat needed a mcoT 10,000 * 4.184x (80-30) 2092000 J so, S 2092 K J اے 4. So, b.) 2 mole is ca@tane produce heat 1.09x10t k5 iso catane require to produce 2092 kJ heet - 2092 1:09x104 = 0.3838532r mole X2 mam moles x molar man isooctane q 0.3838532 X 114.2 43.836 لے

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