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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.

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:

Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from higher-temperature object to lower-temperature object If T1 > T2 q system = - exothermic System (T1) Heat Surroundings (T2) If T1 < T2 q system = + endothermic System (T1) Heat Surroundings (T2)

There are NO temp changes during a phase change.
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 1oC (1 K) Only useable within a state of matter (i.e. s, l, or g) For energy changes involving… heat of fusion (ΔHfus): heat of vaporization (ΔHvap): melting/freezing boiling/condensing There are NO temp changes during a phase change.

Various Specific Heat Capacities
heat capacity (J/K g) Substance Gold Silver Copper Iron Aluminum H2O(l) H2O(s) H2O(g) 0.129 0.235 0.385 0.449 0.897 4.184 2.03 1.998 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

states of matter or during a phase change)
We can find the heat a substance loses or gains using: where q = heat (J) q = m C DT m = mass of substance (g) (used within a given state of matter) C = specific heat (J/goC) DT = temperature change (oC) AND DH = heat of vap/fus (J/g) q = m ΔH Heating Curve HEAT Temp. s s/l l l/g g (used between two states of matter or during a phase change) + Cg ΔHvap Cl ΔHfus Cs D = final – initial

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

Heating Curve of water Gas Boiling Point Liquid Melting Point Solid
140 120 l ↔ g 100 80 Liquid Boiling Point 60 40 Temperature (oC) 20 s ↔ l Melting Point -20 Solid -40 -60 -80 -100 Energy Added

Temperature (oC) Energy Added 140 120 100 80 60 40 20 -20 -40 -60 -80
-20 -40 -60 -80 -100 Energy Added

Heating Curve of Water Temperature (oC) Energy Added 140 120 100 80 60
-20 -40 -60 -80 -100 Energy Added

Heating Curves Temperature Change within phase
change in KE (molecular motion) depends on heat capacity of phase C H2O (l) = J/goC C H2O (s) = J/goC C H2O (g) = J/goC Phase Changes (s ↔ l ↔ g) change in PE (molecular arrangement) temperature remains constant overcoming intermolecular forces (requires the most heat) (requires the least heat) ΔHfus = 333 J/g (s ↔ l) ΔHvap = 2256 J/g (l ↔ g) Why is this so much larger?

Heating Curves Gas - KE  Boiling - PE  Liquid - KE  Melting - PE 
140 Gas - KE  120 100 Boiling - PE  80 60 40 Liquid - KE  Temperature (oC) 20 Melting - PE  -20 -40 Solid - KE  -60 -80 -100 Energy Added

From Ice to Steam in Five Easy Steps
Heating Curve of Water From Ice to Steam in Five Easy Steps q4 q5 q1: Heat the ice to 0°C q2: Melt the ice into a liquid at 0°C q3: Heat the water from 0°C to 100°C q4: Boil the liquid into a gas at 100°C q5: Heat the gas above 100°C q1 = m Cs ΔT q2 = m ΔHfus q3 = m Cl ΔT q4 = m ΔHvap q5 = m Cg ΔT q3 q2 q1 Heat Heat qtot= q1 + q2 + q3 + q4 + q5

Heating Curve Practice
1. How much energy (J) is required to heat 12.5 g of ice at –10.0 oC to water at 0.0 oC? 5 4 3 2 1 Notice that your q values are positive because heat is added… q1: Heat the ice from -10 to 0°C q2: Melt the ice at 0°C to liquid at 0 oC q1 = 12.5 g (2.077 J/g oC)( oC) = q2 = 12.5 g (333 J/g) = J J qtot = q1 + q2 = J + 4,162.5 J = 4,420 J

Heating Curve Practice
2. How much energy (J) is required to heat 25.0 g of ice at –25.0 oC to water at 95.0 oC? Notice that your q values are positive because heat is added… 5 4 3 2 1 q1: Heat the ice from -25 to 0°C q2: Melt the ice at 0°C to liquid at 0 oC q3: Heat the water from 0°C to 95 °C q1 = 25.0 g (2.077 J/g oC)( oC) = q2 = 25.0 g (333 J/g) = J 8325 J q3 = 25.0 g (4.184 J/g oC)(95.0 – 0.0oC) = 9937 J qtot = q1 + q2 + q3 = J + 8,325 J J = 19,560 J

Heating Curve Practice
3. How much energy (J) is removed to cool 50.0 g of steam at oC to ice at -5.0 oC? Notice that your q values are negative because heat is removed… 5 4 3 2 1 q5: Cool the steam from to 100°C q4: Condense the steam into liquid at 100°C q3: Cool the water from 100°C to 0 °C q2: Freeze the water into ice at 0 °C q2 = 50.0 g (- 333 J/g) = q1: Cool the ice from 0°C to – 5.0 °C q5 = 50.0 g (2.042 J/g oC)( oC) = q4 = g ( J/g) = J - -112,800 J q3 = 50.0 g (4.184 J/g oC)(0.0 – 100.0oC) = J J q1 = 50.0 g (2.077 J/g oC)(- 5.0 – 0.0oC) = J qtot = q1 + q2 + q3+ q4 + q5 = J ,800 J J + -16,650 J J = -152,000 J

Food and Energy Caloric Values 1 calorie = 4.184 joules
Food joules/gram calories/gram “Calories”/gram Protein , , Fat , , Carbohydrates 17, , 1 calorie = joules 1000 calories = 1 “Calorie” "science" "food" or… 1 Kcal = 1 “Calorie” Smoot, Smith, Price, Chemistry A Modern Course, 1990, page 51

Does water have negative calories?
How many Calories (nutritional) will you burn by drinking 1.0 L of water, initially at 36.5 oF (standard refrigeration temperature)? Assume that the body must expend energy to heat the water to body temperature at 98.6 oF. 37 oC 1 L = 1000 mL 2.5 oC 1 mL = 1 g 1 calorie = joules 1000 calories = 1 “Calorie” q = 1.0 x 103 g (4.184 J/g oC)(37 oC oC) = 144,348 J J 1 cal 1 “Cal” = 35 Cal 4.184 J 1000 cal

C H2O (l) = 4.184 J/goC C H2O (s) = 2.03 J/goC C H2O (g) = 1.998 J/goC
Heating Curve of H2O Constants and Graph C H2O (l) = J/goC C H2O (s) = 2.03 J/goC C H2O (g) = J/goC ΔHfusion = 6.02 kJ/mol ΔHvap = 40.7 kJ/mol 140 120 100 80 60 40 Temperature (oC) 20 -20 -40 -60 -80 -100 Energy Added 17

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

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

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

(i.e. atmospheric pressure)
Thermometer Styrofoam cover 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 qx = – qy

Heat Transfer Experiments
1. A 75.0 g piece of lead (specific heat = J/goC), initially at 435oC, is set into g of water, initially at 23.0oC. What is the final temperature of the mixture? Pb 125 g 23.0 °C What is the final temperature, Tf, of the mixture? 75.0 g 435.0 °C C = J/°C g qwater = –qPb q = m x C x ΔT for both cases, although specific values differ Plug in known information for each side Solve for Tf ...

A 75.0 g piece of lead (specific heat = 0.130 J/goC),
initially at 435oC, is set into g of water, initially at 23.0oC. What is the final temperature of the mixture? q = m x C x ΔT for both cases, although specific values differ Plug in known information for each side qwater = –qPb mwater Cwater DTwater = –mPb CPb DTPb 125 (4.18) (Tf – 23) = –75 (0.13) (Tf – 435) 522.5 Tf – = –9.75 Tf +9.75 Tf +9.75 Tf Tf = Tf = 30.5oC

2. A 97.0 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15.0oC. If gold has a specific heat of J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter. T = 785oC Au mass = 97.0 g T = 15.0 oC mass = 323 g LOSE heat = GAIN heat - - [(C Au) (mass) (DT)] = (C H2O) (mass) (DT) - [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] - [(12.5) (Tf - 785oC)] = (1.35 x 103) (Tf - 15oC)] -12.5 Tf x = x 103 Tf x 104 3 x 104 = x 103 Tf Tf = oC

HW #2. If 59. 0 g of water at 13. 0 oC are mixed with 87
HW #2. If 59.0 g of water at 13.0 oC are mixed with 87.0 g of water at 72.0 oC, find the final temperature of the system. T = 13.0 oC mass = 59.0 g T = 72.0 oC mass = 87.0 g LOSE heat = GAIN heat - - [ (mass) (C H2O) (DT)] = (mass) (C H2O) (DT) - [ (59 g) (4.184 J/goC) (Tf - 13oC)] = (87 g) (4.184 J/goC) (Tf - 72oC)] - [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)] Tf = Tf = Tf Tf = oC

HW #4. 240. g of water (initially at 20
HW # g of water (initially at 20.0oC) are mixed with an unknown mass of iron initially at 500.0oC (CFe = J/goC). When thermal equilibrium is reached, the mixture has a temperature of 42.0oC. Find the mass of the iron. T = 500oC Fe mass = ? grams T = 20oC mass = 240 g - LOSE heat = GAIN heat -q1 = q2 - [ (mass) (CFe ) (DT)] = (mass) (CH2O) (DT) - [ (X g) ( J/goC) (42oC - 500oC)] = (240 g) (4.184 J/goC) (42oC - 20oC)] - [ (X) (0.4495) (-458)] = (240 g) (4.184) (22) 205.9 X = X = 107 g Fe

KEY: Assume that the ice melts and the final product is a liquid.
A 23.6 g ice cube at –31.0oC is dropped into 98.2 g of water at 84.7oC. Find the equilibrium temperature. KEY: Assume that the ice melts and the final product is a liquid. qice = –qwater qwater = –98.2 (4.18) (Tf – 84.7) = – Tf qice = 23.6 (2.077) (0 – –31) (333) (4.18) (Tf – 0) = Tf = Tf Tf = Tf = 49.9oC

Heating Curve Challenge Problems
Temperature (oC) 40 20 -20 -40 -60 -80 -100 120 100 80 60 140 Time DH = mol x DHfus DH = mol x DHvap Heat = mass x Dt x Cp, liquid Heat = mass x Dt x Cp, gas Heat = mass x Dt x Cp, solid 1. A sample of ice at -25oC is placed into 75 g of water initally at 85oC. If the final temperature of the mixture is 15oC, what was the mass of the ice? 52.8 g ice A 38 g sample of ice at -5oC is placed into 250 g of water at 65oC. Find the final temperature of the mixture assuming that the ice sample completely melts. A 35 g sample of steam at 116oC are bubbled into 300 g water at 10oC. Find the final temperature of the system, assuming that the steam condenses into liquid water. 45.6 oC 76.6 oC

Heating Curve for Water (Phase Diagram)
140 120 100 80 60 40 20 -20 -40 -60 -80 -100 F q4 = m DHvap DHvap = +/ J/g 5 BP q2 = m DHfus D E DHfus = +/- 333 J/g 4 q5 = m C DT C g = J/goC 3 q3 = m C DT Temperature (oC) Cl = J/goC MP B C 2 1 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 q1 = m C DT Cs = J/goC A Heat

Calculating Energy Changes - Heating Curve for Water
140 DH = mol x DHvap 120 DH = mol x DHfus 100 80 60 Heat = mass x Dt x Cp, gas 40 Temperature (oC) 20 Heat = mass x Dt x Cp, liquid -20 -40 -60 Heat = mass x Dt x Cp, solid -80 -100 Time

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