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Energy Changes in Water: Heating and Cooling Curves, Phase Changes, and Calculations

This resource explores the energy changes associated with heating and cooling water, including phase diagrams and calculations of specific heat. It delves into the energy required to heat 25g of liquid water from 25°C to 100°C and the energy released when water cools. The principles of heat of fusion and heat of vaporization are discussed alongside relevant calculations, including the conversion of calories to joules. Practice problems enhance understanding of these concepts, culminating in assignments due on March 29 and 30.

jenibelle-perez
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Energy Changes in Water: Heating and Cooling Curves, Phase Changes, and Calculations

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  1. Calculating energy changes

  2. Heating curve of water

  3. Cooling Curve of Water

  4. Phase diagram

  5. Energy (heat) may be expressed in joules or calories. 1 calorie (cal) = 4.184 joules (J) How many joules in 60.1 calories? How many calories in 28.4 J? Energy conversions

  6. How much energy is required to heat 25 g of liquid water from 25oC to 100oC? Specific heat capacity of liquid water is 4.18 J/goC. Q = mcΔT Q (energy) = m(mass) x c(specific heat) x ΔT (change in temperature = Tf – Ti) Q = (25 g)(4.18 J/goC)(100o – 25oC) If ΔT is positive, Q will be positive. If ΔT is negative, Q will be negative. Calculating Energy Changes

  7. How many joules of heat are given off when 5.0 g of water cools from 75oCto 25oc? (Specific heat of water = 4.18 J/goC)

  8. Page 329 Practice Problem 10.2 Page 330 Practice Problem 10.3 Page 333 Section 10.2 Review Questions 2-6 Your assignment: due Mon.3/29

  9. Energy is increasing/decreasing during phase changes even though temp. remains constant. Q = mass(m) x heat of fusion (Hf) for melting and freezing Q = mass(m) x heat of vaporization (Hv) for boiling and condensing Values will be positive for melting and boiling Values will be negative for freezing and condensing Energy during phase change

  10. How many calories are given off when 85 g of steam condense to liquid water? (Hv= 539.4 cal/g)

  11. How many joules does it take to melt 35g of ice at 0oC? (Hf = 333 J/g)

  12. How many joules are required to convert 10.0g of ice at -10.0oC to steam at 150. oC? • Given: Specific heat of • Specific heat of water = 4.184 J/g C° • Specific heat of Steam = 2.03 J/g C° • Specific heat of ice = 2.06 J/g C° • Hv water = 539.4 cal/g • Hf water = 333.0 J/g • 5 part calculation

  13. Page 497 Practice Problem 14.2 Page 497 Section 14.1 Review 1, 3, 4-7 Page 503 Section 14.2 Review 1, 4, 6 Your assignment: Due tues. 3/30

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