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Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from higher-temperature object to lower-temperature object.

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Presentation on theme: "Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from higher-temperature object to lower-temperature object."— Presentation transcript:

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2 Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from higher-temperature object to lower-temperature object System (T 1 ) Surroundings (T 2 ) Heat If T 1 > T 2 q system = - exothermic System (T 1 ) Surroundings (T 2 ) Heat If T 1 < T 2 q system = + endothermic

3 Calorimetry: the measurement of heat flow device used is called a... calorimeter specific heat capacity (C): amt. of heat needed to raise temp. of 1 g of a substance 1 o C (1 K) Only useable within a state of matter (i.e. s, l, or g) heat of fusion (ΔH fus ): heat of vaporization (ΔH vap ): For energy changes involving… melting/freezing boiling/condensing There are NO temp changes during a phase change.

4 Various Specific Heat Capacities Substance Specific heat capacity (J/K g) Gold Silver Copper Iron Aluminum H 2 O (l) H 2 O (s) H 2 O (g) Metals do not generally require much energy to heat them up (i.e. they heat up easily) Water requires much more energy to heat up

5 We can find the heat a substance loses or gains using: where q = heat (J) m = mass of substance (g) C = specific heat (J/g o C)  T = temperature change ( o C) q = m C  T (used within a given state of matter) AND q = m ΔH (used between two states of matter or during a phase change) HEAT Temp. s s/ l l l /g g Heating Curve ΔH fus ΔH vap C s C l C g + –  = final – initial  H = heat of vap/fus (J/g)

6 q = m  C  ΔT q (J) = mass (g)  C (J/g o C)  ΔT ( o C) q = joules (J) Mnemonic device: q = m “CAT” Using heat capacities…

7 Heating Curve of water Temperature ( o C) Energy Added Melting Point Boiling Point Solid Liquid Gas s ↔ l l ↔ g

8 Temperature ( o C) Energy Added

9 Heating Curve of Water Temperature ( o C) Energy Added

10 Heating Curves Temperature Change within phase change in KE (molecular motion) depends on heat capacity of phase C H 2 O (l) = J/g o C C H 2 O (s) = J/g o C C H 2 O (g) = J/g o C Phase Changes (s ↔ l ↔ g) change in PE (molecular arrangement) temperature remains constant overcoming intermolecular forces (requires the most heat) (requires the least heat) ΔH fus = 333 J/g ΔH vap = 2256 J/g Why is this so much larger? ( l ↔ g) (s ↔ l )

11 Heating Curves Temperature ( o C) Energy Added Melting - PE  Solid - KE  Liquid - KE  Boiling - PE  Gas - KE 

12 Heat q 1 : Heat the ice to 0°C q 2 : Melt the ice into a liquid at 0°C q 3 : Heat the water from 0°C to 100°C q 4 : Boil the liquid into a gas at 100°C q 5 : Heat the gas above 100°C Heating Curve of Water From Ice to Steam in Five Easy Steps q 1 = m C s ΔT q 2 = m ΔH fus q 3 = m C l ΔT q 4 = m ΔH vap q 5 = m C g ΔT q1q1 q2q2 q3q3 q4q4 q5q5 q tot = q 1 + q 2 + q 3 + q 4 + q 5

13 Heating Curve Practice 1. How much energy (J) is required to heat 12.5 g of ice at –10.0 o C to water at 0.0 o C? 4,420 J q 1 : Heat the ice from -10 to 0°C q 2 : Melt the ice at 0°C to liquid at 0 o C q 1 = 12.5 g (2.077 J/g o C)( o C) = q 2 = 12.5 g (333 J/g) = q tot = q 1 + q J = J + 4,162.5 J = J Notice that your q values are positive because heat is added…

14 Heating Curve Practice 2. How much energy (J) is required to heat 25.0 g of ice at –25.0 o C to water at 95.0 o C? 19,560 J q 1 : Heat the ice from -25 to 0°C q 2 : Melt the ice at 0°C to liquid at 0 o C q 3 : Heat the water from 0°C to 95 °C q 1 = 25.0 g (2.077 J/g o C)( o C) = q 2 = 25.0 g (333 J/g) = q tot = q 1 + q 2 + q J 8325 J = J + 8,325 J J = q 3 = 25.0 g (4.184 J/g o C)(95.0 – 0.0 o C) = 9937 J Notice that your q values are positive because heat is added…

15 Heating Curve Practice 3. How much energy (J) is removed to cool 50.0 g of steam at o C to ice at -5.0 o C? -152,000 J q 5 : Cool the steam from to 100°C q 4 : Condense the steam into liquid at 100°C q 3 : Cool the water from 100°C to 0 °C q 2 : Freeze the water into ice at 0 °C q 2 = 50.0 g (- 333 J/g) = q 1 : Cool the ice from 0°C to – 5.0 °C q 5 = 50.0 g (2.042 J/g o C)( o C) = q 4 = 50.0 g ( 2256 J/g) = q tot = q 1 + q 2 + q 3 + q 4 + q J -112,800 J = J ,800 J J + -16,650 J J = q 3 = 50.0 g (4.184 J/g o C)(0.0 – o C) = J J q 1 = 50.0 g (2.077 J/g o C)(- 5.0 – 0.0 o C) = J Notice that your q values are negative because heat is removed… -

16 Food and Energy Caloric Values Food joules/gram calories/gram “Calories”/gram Protein 17,000 4,000 4 Fat 38,000 9,000 9 Carbohydrates 17,000 4,000 4 Smoot, Smith, Price, Chemistry A Modern Course, 1990, page calories = 1 “Calorie” "science" "food" 1 calorie = joules or… 1 Kcal = 1 “Calorie”

17 Does water have negative calories? How many Calories (nutritional) will you burn by drinking 1.0 L of water, initially at 36.5 o F (standard refrigeration temperature)? Assume that the body must expend energy to heat the water to body temperature at 98.6 o F. 37 o C 2.5 o C 1 L = 1000 mL 1 mL = 1 g q = 1.0 x 10 3 g (4.184 J/g o C)(37 o C o C) = 144,348 J 1000 calories = 1 “Calorie” 1 calorie = joules J1 cal J = 35 Cal 1 “Cal” 1000 cal

18 C H 2 O (l) = J/g o C C H 2 O (s) = 2.03 J/g o C C H 2 O (g) = J/g o C Temperature ( o C) Energy Added ΔH fusion = 6.02 kJ/mol ΔH vap = 40.7 kJ/mol Heating Curve of H 2 O Constants and Graph

19 What will happen over time? Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 291

20 Let’s take a closer look… Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 291

21 Eventually, the temperatures will equalize Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page 291

22 Thermometer Styrofoam cover Styrofoam cups Stirrer Much calorimetry is carried out using a coffee-cup calorimeter, under constant pressure (i.e. atmospheric pressure) If we assume that no heat is lost to the surroundings, then the energy absorbed inside the calorimeter must be equal to the energy released inside the calorimeter. i.e., q absorbed = – q released qxqx = – q y

23 125 g 23.0 °C Heat Transfer Experiments q water = –q Pb q = m x C x ΔT for both cases, although specific values differ Plug in known information for each side Solve for T f... Pb 75.0 g °C C = J/°C g What is the final temperature, T f, of the mixture? 1. A 75.0 g piece of lead (specific heat = J/g o C), initially at 435 o C, is set into g of water, initially at 23.0 o C. What is the final temperature of the mixture?

24 A 75.0 g piece of lead (specific heat = J/g o C), initially at 435 o C, is set into g of water, initially at 23.0 o C. What is the final temperature of the mixture? m water C water  T water q water = –q Pb –m Pb C Pb  T Pb = 125 (4.18) (T f – 23) = T f – =–9.75 T f T f T f = T f = 30.5 o C –75 (0.13) (T f – 435) q = m x C x ΔT for both cases, although specific values differ Plug in known information for each side

25 2. A 97.0 g sample of gold at 785 o C is dropped into 323 g of water, which has an initial temperature of 15.0 o C. If gold has a specific heat of J/g o C, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. T = 785 o C mass = 97.0 g T = 15.0 o C mass = 323 g LOSE heat = GAIN heat - - [(C Au) (mass) (  T)] = (C H 2 O) (mass) (  T) - [(0.129 J/g o C) (97 g) (T f o C)] = (4.184 J/g o C) (323 g) (T f - 15 o C)] - [(12.5) (T f o C)] = (1.35 x 10 3 ) (T f - 15 o C)] T f x 10 3 = 1.35 x 10 3 T f x x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C Au

26 HW #2. If 59.0 g of water at 13.0 o C are mixed with 87.0 g of water at 72.0 o C, find the final temperature of the system. T = 13.0 o C mass = 59.0 g LOSE heat = GAIN heat - - [ (mass) (C H 2 O) (  T)] = (mass) (C H 2 O) (  T) - [ (59 g) (4.184 J/g o C) (T f - 13 o C)] = (87 g) (4.184 J/g o C) (T f - 72 o C)] - [(246.8) (T f - 13 o C)] = (364.0) (T f - 72 o C)] T f = 364 T f = T f T f = 48.2 o C T = 72.0 o C mass = 87.0 g

27 HW # g of water (initially at 20.0 o C) are mixed with an unknown mass of iron initially at o C (C Fe = J/g o C). When thermal equilibrium is reached, the mixture has a temperature of 42.0 o C. Find the mass of the iron. T = 500 o C mass = ? grams T = 20 o C mass = 240 g LOSE heat = GAIN heat - - [ (mass) (C Fe ) (  T)] = (mass) (C H2O ) (  T) - [ (X g) ( J/g o C) (42 o C o C)] = (240 g) (4.184 J/g o C) (42 o C - 20 o C)] - [ (X) (0.4495) (-458)] = (240 g) (4.184) (22) X = X = 107 g Fe -q 1 = q 2 Fe

28 A 23.6 g ice cube at –31.0 o C is dropped into 98.2 g of water at 84.7 o C. Find the equilibrium temperature. KEY: Assume that the ice melts and the final product is a liquid. q ice = –q water q ice = 23.6 (2.077) (0 – –31) (333) (4.18) (T f – 0) T f = T f = 49.9 o C q water = –98.2 (4.18) (T f – 84.7) = T f = T f = – T f

29 Heating Curve Challenge Problems 1. A sample of ice at -25 o C is placed into 75 g of water initally at 85 o C. If the final temperature of the mixture is 15 o C, what was the mass of the ice? Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid 2.A 38 g sample of ice at -5 o C is placed into 250 g of water at 65 o C. Find the final temperature of the mixture assuming that the ice sample completely melts. 3.A 35 g sample of steam at 116 o C are bubbled into 300 g water at 10 o C. Find the final temperature of the system, assuming that the steam condenses into liquid water g ice 45.6 o C 76.6 o C

30 A  B warm ice B  C melt ice (s  l) C  D warm water D  E boil water (l  g) E  D condense steam (g  l) E  F superheat steam Heating Curve for Water (Phase Diagram) Temperature ( o C) Heat BP MP A B C D E F q 2 = m  H fus  H fus = +/- 333 J/g q 4 = m  H vap  H vap = +/ J/g q 3 = m C  T C l = J/g o C q 1 = m C  T C s = J/g o C q 5 = m C  T C g = J/g o C

31 Calculating Energy Changes - Heating Curve for Water Temperature ( o C) Time  H = mol x  H fus  H = mol x  H vap Heat = mass x  t x C p, liquid Heat = mass x  t x C p, gas Heat = mass x  t x C p, solid


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