 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.

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Presentation transcript:

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

 How much energy is required to heat 25 g of liquid water from 25 o C to 100 o C? Specific heat capacity of liquid water is 4.18 J/g o C.  Q = mcΔT  Q (energy) = m(mass) x c(specific heat) x ΔT (change in temperature = T f – T i )  Q = (25 g)(4.18 J/g o C)(100 o – 25 o C)  If ΔT is positive, Q will be positive. If ΔT is negative, Q will be negative.

 How many joules of heat are given off when 5.0 g of water cools from 75 o C to 25 o c? (Specific heat of water = 4.18 J/g o C )

 Page 329 Practice Problem 10.2  Page 330 Practice Problem 10.3  Page 333 Section 10.2 Review Questions 2-6

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

 How many calories are given off when 85 g of steam condense to liquid water? (H v = cal/g)

 How many joules does it take to melt 35g of ice at 0 o C? (H f = 333 J/g)

 How many joules are required to convert 10.0g of ice at o C to steam at 150. o C?  Given: Specific heat of  Specific heat of water = J/g C°  Specific heat of Steam = 2.03 J/g C°  Specific heat of ice = 2.06 J/g C°  H v water = cal/g  H f water = J/g  5 part calculation

 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