Heating Cooling Curves Worksheet Answers

6. Energy and Chemical Reactions HEATING CURVE CALCULATIONS Temperature (°C) 100 50 Time -> In the heating and cooling curves we learned that energy is absorbed by a substance as it warms up, melts (fusion) or boils (vaporization) and energy is released from a substance as it cools down, condenses, or freezes. Calorimetry (q=mCAT) allows us to calculate the energy changes as a substance warms or cools. (1.3. & 5) The energies involved in phase changes (arcas 2 & 4) are the Heat of Vaporization (liquid gas) and the Heat of Fusion (solid liquid). These energies will be used as conversion factors. Heat of Vaporization or Heat of Condensation of water Heat of Fusion (melting) or Heat of Solidification of water Huap 23301 gram H 335 J gram Joules (J) are energy units. It takes 4,184 Joules of energy to beat 1 gram of water by 1 °C. Examples! Calculate the energy needed to vaporize 10.0 g of water. 10.0 g H2O x 23301 gram 23,000 J23.0 kJ Calculate the energy released when 10.0 kg of water melts. 10.0 kg H₂O x 1000g 335 J 1 kg X gram 3,350,000 13,350 kl Do the following calculations. Show your equation for cach problem. Box your answers. 135.0k3 SF. 1. Calculate the energy needed to vaporize... a) 15.0 g of water 25303 34,9505 34 34.95 kJ T b) 5.75 kg of water 1000g 2330 ikg 13,397,5005 13,400 kJ 3.88 moles of water 18.029 23303=1162,9083 163 45 2. Calculate the mass of water (in grams) that will be vaporized by.... 1) 20.0 kJ of energy 1000312 143 23303 8.582 b) 175 kJ of energy 1000g 23305 A 75.12 c) 135 J of energy 23300 6.6379 2 3. Calculate the energy needed to melt... a) 23.0 g of water v 335.3 7,705 J = 7,710.J-7.71kJ 12 b) 8.75 kg of water 1000g 33.5J 2,131,2305 = 2,930 kJ c) 3.25 moles of water 3353 19,619 J= 19.6 KJ Calculate the mass of water (in grans) that will be melted by... 89.559 = 89.48 a) 18003 30.0 kJ of energyTE3 ४ 3353 C d) absorbed as 13.6 g of water vaporizes 2330 31,688 JA 31.7 kJ b) 7.60 kJ of energy 10003 143 3353 22. c) 133 J of energy 3353 0.397g 5. Calculate the energy... absorbed by 35.8 g of ice moting 3353, 11,993 3 = ง 12.0k b) released as 88.5 g of water vapor condenses 23305 206, 205J= 206KJ 2 released as 92.2 g of water freezes 3357 2 30,887 3:30.9 kJ e) absorbed when 2.25 moles of ice melts 10 18.625 3355 13,58245=13.6kJ 1) absorbed when 2.25 moles of water vaporizes 18.02 23303 = 94,469 J = 94.5KJ 6. A 25.00 gram sample of ice at 0.00°C melts and then warms up to 20.0°C. How much energy is absorbed? TWO STE MELTS 335.J8,375 J 25.00gv8 WARMS 8 (25.00%) (4.1845) (20.0%) = 2,0923 From Summer! Toom: 10,467 5:10.47kJ