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Name Date Pd Unit 3 Worksheet 4 – Quantitative Energy Problems Part 2 Energy constants (H2O) Q m c T(C) Heat of fusion (melting or freezing) Hf Q m H v or m H f Heat of vaporization (evaporating or condensing) Hv Heat capacity (c) of solid water Heat capacity (c) of liquid water For each of the problems sketch a warming or cooling curve to help you decide which equation(s) to use to solve the problem. Keep a reasonable number of sig figs in your answers. 334 J/g 2260 J/g 2.1 J/g˚C 4.18 J/g˚C T (°C) 1. How much energy must be absorbed by a 150 g sample of ice at 0.0 ˚C that melts and then warms to 25.0˚C? 2) liquid H2O warms up 25 1) ice melts -25 Q1 m H f Q2 m c T Q1 150 g 334 J g Q1 50,100 J Q2 150 g 4.18 J gC 25C Q2 15,700 J QTotal Q1 Q2 50,100J 15,700J 65,800J 66,000J 66kJ time (min) T (°C) 2. Suppose in the Icy Hot lab that the burner transfers 325 kJ of energy to 450 g of liquid water at would be boiled away? 20.˚C. What mass of the water 2) some H2O boils 100 50 1) H2O warms to 100°C 0 tim e (m in) Q1 m c T Q1 450g 4.18 J gC 80C Q1 150kJ Q2 QT Q1 Q2 325kJ 150kJ Q2 m H v Q m 2 Hv 175,000J m 77.4g 77g 2260 J g Q2 175kJ 3. A 12oz can of soft drink (assume m = 340 g) at 25˚C is placed in a freezer where the temperature is – 12 ˚C. How much energy must be removed from the soft drink for it to reach this temperature? T (°C) 25 1 2 Q1 m c T 3 Q1 340 g 4.18 J -25 tim e (m in) 3 steps: 1) cool H2O to 0°C 2) freeze H2O 3) cool ice liquid : ©Modeling Instruction – AMTA 2012 freezing H 2O : Q2 m H f 25C gC Q1 35,5300 J or 35.5 kJ solid : Q2 340 g 334 J g Q2 113,560 J or 113.6 kJ Q3 m c T 35.5 kJ 113 .6 kJ 8.6 kJ 158 .6 kJ or 160 kJ Q3 340 g 2.1J /gC 12C Q3 8568 J or 8.6 kJ 1 QTotal Q1 Q2 Q3 U3 ws4 v3.0 ©Modeling Instruction – AMTA 2012 2 U3 ws4 v3.0 4. 65.0 kilojoules of energy are added to 150 g of ice at 0.0˚C. What is the final temperature of the water? T (°C) 30 ? °C 2 10 1 final temperature : melting ice : Q1 m H f Q2 m c T Q1 150 g 334 J / g T Q2 QTotal Q1 Q2 mc 14,900 J T 150 g 4.18 J / g C Q2 65.0 kJ 50.1 kJ 14.9 kJ T 23.8C Q1 50.1 k J -10 tim e (m in) Q2 14,900 J 5. 250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze? Explain. maximum mass : Q2 m H f Q1 m c T Q1 400 g 4.18 J / g C 60C T (°C) 65 40 1 m Q1 100 kJ 15 2 Q2 QTotal Q1 -10 Q2 250 kJ 100 kJ 150 kJ tim e (m in) Q2 150,000 J 449 g Hf 334 J / g Our mass is less than the maximum mass 250kJ could freeze so we could freeze even more than the 400g sample. Q2 150,000 J 6. An ice cube tray full of ice (235g) at –7.0˚C is allowed to warm up to room temperature (22˚C). 1 How much energy must be absorbed by the contents of the tray in order for this to happen? Q1 m c T melting ice : Q2 m H f Q1 235 g 2.10 J / g C 7C Q 2 235 g 334 J / g Q1 3.5 kJ Q 2 78.5 k J liquid : Q3 m c T QTotal Q1 Q2 Q3 T (°C) solid : 15 3 2 1 -10 tim e (m in) 3.5 kJ 78.5 kJ 21.6 kJ 104 kJ Q3 235 g 4.18 J / g C 22C Q3 21.6 kJ If this same quantity of energy were removed from 40.0 g of water vapor at 100˚C, what would be the final temperature of the water? T (°C) 7. 125 100 75 50 25 0 1 2 ? °C tim e (m in) condensing vapor: Q1 m H v change in temperature : Q2 m c T Q1 40.0 g 2260 J / g T Q1 90.4 kJ Q2 13,600 J m c 40.0 g 4.18 J / g C T 81.3C Q2 QTotal Q1 Q2 104 kJ 90.4 kJ 13.6 kJ T final 100C 81.3C 18.7C Q2 13,600 J ©Modeling Instruction – AMTA 2012 3 U3 ws4 v3.0