Specific Heat

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Specific Heat
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Different substances have different abilities to
store energy. They are said to have different
heat capacities.
Heat capacity is defined as the amount of heat
required to change the temperature of the
substance by one degree Celsius (or Kelvin).
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Most metals have low specific heats, while
nonmetal compounds mixtures such as water,
wood, soil, air have relatively high specific
heats.
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Calorimeter

• A device that measures temperature changes in
  surroundings
• Heat transferred by physical and chemical changes
  can be measured using a process called
  calorimetry.

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Chem Saver p 41CALORIMETRY

• Calorimetry Process for determining the amount
  of heat energy released or absorbed in a chemical
  or physical change.
•  
• Enthalpy Change in Heat energy Symbolized by H
• Entropy extensive property which measures the
  degree of disorder.

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Units of heat include

• calorie - the amount of heat required to change
  the temperature of 1 gram of pure liquid water by
  one degree Celsius.
• Food calorie (Calorie, big calorie) - is equal to
  1000 calories or one kilocalorie
• Joule - International System unit of energy.
  There are 4.18 Joules in one calorie.
• Kilojoule - 1000 joules
• For heat conversions use
• 1Kcalorie 1Calorie 1000 calories
• 1calorie 4.18J

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Foods and Heat Energy

• 1 nutritional Calorie, 1 Cal 1000 cal 1 kcal.
• Energy in our bodies comes from carbohydrates and
  fats (mostly).
• Intestines carbohydrates converted into glucose
• C6H12O6 6O2 ? 6CO2 6H2O, DH -2816 kJ
• Fats break down as follows
• 2C57H110O6 163O2 ? 114CO2 110H2O, DH
  -75,520 kJ
• Fats contain more energy are not water soluble,
  so are good for energy storage.

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A popscicle has 60.0 Calories per serving. How
many calories is this?
Problem 1

• 1Kcalorie 1Calorie 1000 calories
• 1calorie 4.18J
• 60.0Calories x 1000calories 60,000 cal or
  
• 1 1Cal
  6.00x104cal

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Amazing fruit candy has 10.0 Calories per
serving. How many joules is this?
Problem 2

• 1Kcalorie 1Calorie 1000 calories
• 1calorie 4.18J
• 10.0 Calories x 1000 cal x 4.18J 41800J
• 1 1Cal 1cal

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Chem Saver Page 41 Calculating Heat Energy

• Since the calorie is defined in terms of water,
  the heat capacity for liquid water is 1 cal/g oC.
  
• This also equates to 4.18 J/g oC.
• Calculating calorimetry problems
• Q m x Cp x ?T
• Q heat (cal or J)
• m mass of the substance (g)
• Cp heat capacity (cal/g oC or J/g oC)
• and D T change in temperature of the substance
  (oC)

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Heat Capacity

• The greater the mass of the object, the greater
  its heat capacity.
• A massive steel cable on a bridge requires much
  more heat to raise its temperature 1ºC than a
  small steel nail does.
• Different substances with the same mass may have
  different heat capacities.
• On a sunny day, a 20-kg puddle of water may be
  cool, while a nearby 20-kg iron sewer cover may
  be too hot to touch.

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Problem 7

• When a hamburger is burned in a calorimeter,
  2000. g of water increases in temperature by 30.0
  oC. How many Calories are in the hamburger?
• Q(m) (Cp) (?T)

• Cwater 1 cal/g oC or 4.18 J/g oC
• ?T 30 ?C

• Q 2000. g x 1 cal/g oC x 30.0 oC 60000cal
• 60000cal x 1Cal 60.0 Cal
• 1 1000cal

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Problem 9

• The temperature of 2500 grams of mercury rises
  from 20 oC to 60oC when it absorbs 13,794 joules
  of heat. Calculate the specific heat capacity of
  the mercury.
• Q(m) (Cp) (?T)

• Cmercury ?
• ?T 40 oC

• 13,794 J 2500g x (Cp) x 40.0 oC
• 13,794 J 100000g oC x (Cmercury)
• (Cmercury) 0.138 J/g oC

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Problem 11a

• An 800-gram block of lead is heated in boiling
  water (100 oC) until its temperature is the same
  as the boiling water. The lead is then removed
  from the boiling water and dropped into 250 grams
  of cool water at 12.2 oC. After a short time,
  the temperatures of both lead and water levels
  out at 20.0 oC.
• Calculate the amount of heat (in Joules) gained
  by the cool water.
• Q(m) (Cp) (?T)

m 250g Ti 12.2 ?C and Tf 20.0 ?C ?T 7.8 ?C
Cwater 4.18 J/g?C
Q (250g) ( 4.18 J/g?C) ( 7.8 ?C ) Q 8151J
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Problem 11b

• An 800-gram block of lead is heated in boiling
  water (100 oC) until its temperature is the same
  as the boiling water. The lead is then removed
  from the boiling water and dropped into 250 grams
  of cool water at 12.2 oC. After a short time,
  the temperatures of both lead and water levels
  out at 20.0 oC.
• Calculate the specific heat capacity of the lead
  based on these measurements, assuming that no
  heat was lost in the process.
• Q(m) (Cp) (?T)

m 800g ?T 80 ?C CPb ? Q gained by water Q
lost by Pb 8151J
8151J (800g) (CPb) (80 ?C) CPb 0.127 J/g?C
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Energy and Change of State

• At the freezing or boiling point two phases of
  matter can exist at the same temperature
• To make the change from one phase to another more
  energy will be absorbed (boiling or melting) or
  lost (condensing or freezing) without a change in
  temperature
• This is because this energy is used merely to
  overcome the bonds of one state and move to the
  new state creating a change in potential energy.

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Chem Saver page 16Heating and Cooling Curve
Heat of vaporization
Gas/vapor
D PE
Boiling/vaporization
D KE
condensation
Heat of fusion
D PE
liquid
melting
D KE
D KE
Solid
freezing
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Heat in phase changes

• Instead of specific heat (C) we use enthalpy (?H)
  for calculating heat during phase changes.
• Heat of fusion/solidification is the heat
  required to move from solid ?? liquid
• Hfus Hsolid
• Heat of vaporization/condensation is the heat
  required to move from liquid ?? gas
• Hvap Hcon

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Heat in phase changes

• Q m x H

• Q amount of heat energy (joules or calories)
• m mass of substance (grams)
• H enthalpy of fusion (Hf) or vaporization (Hv)

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Chemistry in ActivityChem Saver Page 41

• constants
• Specific heat of ice 2.09 J/gºC
• Specific heat of water 4.18 J/gºC
• Specific heat of steam 2.03 J/gºC
• Heat of fusion of water 334 J/g
• Heat of vaporization of water 540 J/g

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Example Problem

• Calculate the mass of ice (in grams) that will
  melt at 0ºC if 2.25 kJ of heat are added. (Hf
  334 J/g)
• Q m x Hf

• Q 2250 J
• Hf 334 J/g
• m Q /Hf
• m 2250 J / 334 J/g

m 6.74 g
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Example Problem

• Calculate the mass of water vapor (in grams) at
  100ºC that can be condensed into liquid at 100ºC
  if 55.0 kJ of heat is removed. (Hv 2257 J/g)
• Q m x Hv

• Q 55000 J
• Hv 2257 J/g m
• m Q /Hv
• m 55000 J / 2257 J/g
• m 24.4 g

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Sample Problem

• How much heat does it take to turn a 20 g chunk
  of ice at -40oC into 20 g of steam at 120oC?
• This is a 5 step problem, each segment of the
  graph must be calculated separately and then
  added together to get a total heat absorbed.

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Step 1
-40oC
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Step 1

• How much heat does it take to turn a 20 g chunk
  of ice at -40oC into 20 g of ice at 0oC?
• Q(m) (Cp) (?T)

m 20g Ti -40 ?C and Tf 0 ?C ?T Tf - Ti 0
?C (-40.0 ?C) 40 ?C Cice 2.09 J/gºC
Q (20g) ( 2.09 J/g?C) ( 40 ?C ) Q 1672J (Q
indicates heat gained or endothermic)
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Step 2

• How much heat does it take to turn a 20 g chunk
  of ice at 0oC into 20 g of water at 0oC?
• Q(m) (Hf)

m 20g Hf 334 J/g
Q (20g) ( 334J/g) Q 6680J (Q indicates heat
gained or endothermic)
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Step 3

• How much heat does it take to turn a 20 g water
  at 0oC into 20 g of water at 100oC?
• Q(m) (Cp) (?T)

m 20g Ti 0 ?C and Tf 100 ?C ?T Tf - Ti
100 ?C 0 ?C 100 ?C Cwater 4.18 J/gºC
Q (20g) ( 4.18 J/g?C) ( 100 ?C ) Q 8360J (Q
indicates heat gained or endothermic)
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Step 4

• How much heat does it take to turn a 20 g water
  at 100oC into 20 g of water vapor at 100oC?
• Q(m) (Hv)

m 20g Hf 334 J/g
Q (20g) ( 334J/g) Q 6680J (Q indicates heat
gained or endothermic)
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Entropy

• Entropy is a measure of how chaotic a system is.
  The less order that is present the more entropy.
  In terms of states of matter

Liquid
Gas
Solid
Low Entropy
High Entropy