Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from...
25
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
Heat (q) Heat: the transfer of energy between objects due to a temperature difference Flows from higher-temperature object to lower-temperature object
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
Slide 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 (H fus ): heat of
vaporization (H vap ): For energy changes involving
melting/freezing boiling/condensing There are NO temp changes
during a phase change.
Slide 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) 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
Slide 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 H (used between two states of matter or during a
phase change) HEAT Temp. s s/ l l l /g g Heating Curve H fus H vap
C s C l C g + = final initial H = heat of vap/fus (J/g)
Slide 6
q = m C T q (J) = mass (g) C (J/g o C) T ( o C) q = joules (J)
Mnemonic device: q = m CAT Using heat capacities
Slide 7
Heating Curve of water Temperature ( o C) 40 20 0 -20 -40 -60
-80 -100 120 100 80 60 140 Energy Added Melting Point Boiling Point
Solid Liquid Gas s l l g
Slide 8
Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60
140 Energy Added
Slide 9
Heating Curve of Water Temperature ( o C) 40 20 0 -20 -40 -60
-80 -100 120 100 80 60 140 Energy Added
Slide 10
Heating Curves Temperature Change within phase change in KE
(molecular motion) depends on heat capacity of phase C H 2 O (l) =
4.184 J/g o C C H 2 O (s) = 2.077 J/g o C C H 2 O (g) = 2.042 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) H fus = 333 J/g
H vap = 2256 J/g Why is this so much larger? ( l g) (s l )
Slide 11
Heating Curves Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100
120 100 80 60 140 Energy Added Melting - PE Solid - KE Liquid - KE
Boiling - PE Gas - KE
Slide 12
Heat q 1 : Heat the ice to 0C q 2 : Melt the ice into a liquid
at 0C q 3 : Heat the water from 0C to 100C q 4 : Boil the liquid
into a gas at 100C q 5 : Heat the gas above 100C Heating Curve of
Water From Ice to Steam in Five Easy Steps q 1 = m C s T q 2 = m H
fus q 3 = m C l T q 4 = m H vap q 5 = m C g T q1q1 q2q2 q3q3 q4q4
q5q5 q tot = q 1 + q 2 + q 3 + q 4 + q 5
Slide 13
5 4 3 2 1 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 0C q 2 : Melt the ice at 0C
to liquid at 0 o C q 1 = 12.5 g (2.077 J/g o C)(0.0 - -10.0 o C) =
q 2 = 12.5 g (333 J/g) = q tot = q 1 + q 2 259.63 J = 259.63 J +
4,162.5 J = 4162.5 J Notice that your q values are positive because
heat is added
Slide 14
5 4 3 2 1 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 0C q 2 : Melt the ice at 0C
to liquid at 0 o C q 3 : Heat the water from 0C to 95 C q 1 = 25.0
g (2.077 J/g o C)(0.0 - -25.0 o C) = q 2 = 25.0 g (333 J/g) = q tot
= q 1 + q 2 + q 3 1298.1 J 8325 J = 1298.1 J + 8,325 J + 9937 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
Slide 15
5 4 3 2 1 Heating Curve Practice 3. How much energy (J) is
removed to cool 50.0 g of steam at 115.0 o C to ice at -5.0 o C?
-152,000 J q 5 : Cool the steam from 115.0 to 100C q 4 : Condense
the steam into liquid at 100C q 3 : Cool the water from 100C 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 0C to 5.0 C q 5 = 50.0 g (2.042 J/g o
C)(100.0 - 115.0 o C) = q 4 = 50.0 g ( 2256 J/g) = q tot = q 1 + q
2 + q 3 + q 4 + q 5 -1531.5 J -112,800 J = -1531.5 J + -112,800 J +
-20920 J + -16,650 J + -519.25 J = q 3 = 50.0 g (4.184 J/g o C)(0.0
100.0 o C) = -20920 J -16650 J q 1 = 50.0 g (2.077 J/g o C)(- 5.0
0.0 o C) = -519.25 J Notice that your q values are negative because
heat is removed -
Slide 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 51 1000 calories = 1 Calorie "science"
"food" 1 calorie = 4.184 joules or 1 Kcal = 1 Calorie
Slide 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 - 2.5 o C) = 144,348 J 1000 calories = 1
Calorie 1 calorie = 4.184 joules 144348 J1 cal 4.184 J = 35 Cal 1
Cal 1000 cal
Slide 18
C H 2 O (l) = 4.184 J/g o C C H 2 O (s) = 2.03 J/g o C C H 2 O
(g) = 1.998 J/g o C Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100
120 100 80 60 140 Energy Added H fusion = 6.02 kJ/mol H vap = 40.7
kJ/mol Heating Curve of H 2 O Constants and Graph
Slide 19
What will happen over time? Zumdahl, Zumdahl, DeCoste, World of
Chemistry 2002, page 291
Slide 20
Lets take a closer look Zumdahl, Zumdahl, DeCoste, World of
Chemistry 2002, page 291
Slide 21
Eventually, the temperatures will equalize Zumdahl, Zumdahl,
DeCoste, World of Chemistry 2002, page 291
Slide 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
Slide 23
125 g 23.0 C Heat Transfer Experiments q water = q Pb q = m x C
x T for both cases, although specific values differ Plug in known
information for each side Solve for T f... Pb 75.0 g 435.0 C C =
0.130 J/C g What is the final temperature, T f, of the mixture? 1.
A 75.0 g piece of lead (specific heat = 0.130 J/g o C), initially
at 435 o C, is set into 125.0 g of water, initially at 23.0 o C.
What is the final temperature of the mixture?
Slide 24
A 75.0 g piece of lead (specific heat = 0.130 J/g o C),
initially at 435 o C, is set into 125.0 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) =
522.5 T f 12017.5 =9.75 T f + 4241.25 +9.75 T f +12017.5 532.25 T f
=16258.75 T f = 30.5 o C 75 (0.13) (T f 435) q = m x C x T for both
cases, although specific values differ Plug in known information
for each side
Slide 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 0.129 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 - 785 o C)] = (4.184 J/g o C) (323 g)
(T f - 15 o C)] - [(12.5) (T f - 785 o C)] = (1.35 x 10 3 ) (T f -
15 o C)] -12.5 T f + 9.82 x 10 3 = 1.35 x 10 3 T f - 2.02 x 10 4 3
x 10 4 = 1.36 x 10 3 T f T f = 22.1 o C Au
Slide 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)] -246.8 T f + 3208 = 364 T f - 26208
29416 = 610.8 T f T f = 48.2 o C T = 72.0 o C mass = 87.0 g
Slide 27
HW #4. 240. g of water (initially at 20.0 o C) are mixed with
an unknown mass of iron initially at 500.0 o C (C Fe = 0.4495 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) (0.4495 J/g o
C) (42 o C - 500 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) 205.9 X = 22091 X =
107 g Fe -q 1 = q 2 Fe
Slide 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) + 23.6 (333) + 23.6 (4.18) (T f 0) 509.13 T f
=25388.99 T f = 49.9 o C q water = 98.2 (4.18) (T f 84.7) = 1519.53
+ 7858.8 + 98.65 T f = 9378.33 + 98.65 T f = 410.48 T f +
34767.32
Slide 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) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140
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.
52.8 g ice 45.6 o C 76.6 o C
Slide 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) 140 120 100 80 60 40 20 0 -20 -40 -60 -80
-100 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 = +/- 2256 J/g q 3 = m C T C l =
4.184 J/g o C q 1 = m C T C s = 2.077 J/g o C q 5 = m C T C g =
2.042 J/g o C 1 2 3 4 5
Slide 31
Calculating Energy Changes - Heating Curve for Water
Temperature ( o C) 40 20 0 -20 -40 -60 -80 -100 120 100 80 60 140
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