Question

a) calculate the temperature change for a 987 gram sample of water that absorbed 6,990 joules of heat. is this endo- or exorthermic? b) how many grams of water when absorbing 12,000 calories will heat from 31.2 degrees c to 99.5 degrees c? c) 69.8 grams of aluminum is heated with 3,900 joules of energy. what will be the temperature change? what would the temperature be if the initial temperature was 25.0 degrees c? the c for aluminum is = 0.897 j/g degrees c

Answer 1

Solution &= mx cx oo ga heat absorbed 26990 me mass of water a 987 g (= specific heat - 4.184 Igre 10 = change in temprature 와 - 6990 = 997 X 4.184 x AO Ads 6990 987 X 4.18 4 Endothermic 6990 4129.6 ) 1.69°C. 1 (6) Q = mx cx Do = 12000 cal اول له [ ع AO qq. 5 - 31.2 = 68.3 m2 ? 12000= mx 1x 68.3 so m2 :) 12000 68.3

(2 м х Су до 92 39оо л Д ? (4: Ка С. О. Я 4 (о) 3400 = 6:8xo-C47Xде от 39oo 2 ДО 62-6103 62: 28 ДО об 62: 282 Те - 1; с » Бі: 28 ТТ 42:27+25 е 2. 20°С