Question

y pay tom H2o Oy soon as to, or forved othe 3. Which of the following describes your average heat of neutralization? Why? • Accurate and precise jsoulade . Accurate but not precise does not . Precise but not accurate • Neither accurate nor precise. change halpy of ation. 6. On a hot day, you take a six-pack of soda on a picnic, cooling it with ice. Each empty aluminum can weighs 12.5 g and contains 355 mL of soda. The specific heat of aluminum is 0.902 J/g."C; the specific heat of the soda is 4.10 J/g.ºC; take the density of the soda as equal to that of water (1.00 g/mL). How much heat must be absorbed from the six pack to lower the temperature from 33.3°C to 4.0°C?

Answer 1

9=MCAT 9- heat absorbed / released m = mass (= specific heat capacity IT-change in temperature 9 total - 9 sodat gean Proloco Msode = density x Volume Mean = 12:54 - = 1g/ml x 355ml ton = 355 g Csoda = 4.10519C (can = 0.902 J AT= 33.3°C -4°C -1 AT = 29.3°C - 29.3 °C 9 - 365 g x 4. 105/90 c 9 = 12.59 x 0.902 5/96. X 29.30 X 29.3°C = 42646.15J 330:36 J since there are 6 cans Since there are 6 can of soda 9 e 6x330. 365 92 6.x 4264615J 1982:16J 1 - 255876.9 J 255876.95 + 1982:16 J 2578 59.06 J 257.9 kJ