Section 7.2— Calorimetry & Heat Capacity hy do some things get hot more quickly than others?
Section 7.2—Calorimetry & Heat Capacity
Feb 24, 2016
Section 7.2—Calorimetry & Heat Capacity. Why do some things get hot more quickly than others?. Temperature. Temperature – proportional to the average kinetic energy of the molecules. Energy due to motion. (Related to how fast the molecules are moving). As temperature increases. - PowerPoint PPT Presentation
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Section 7.2—Calorimetry & Heat Capacity
Why do some things get hot more quickly than others?
Temperature
Temperature – proportional to the average kinetic energy of the molecules
Energy due to motion(Related to how fast the molecules are moving)
As temperature increases
Molecules move faster
Heat & Enthalpy
Heat (q)– The flow of energy from higher temperature particles to lower temperature particles
Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”
Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume
Energy Units
The most common energy units are Joules (J) and calories (cal)
4.18 J 1.00 cal
1000 J
1000 cal
1 kJ
1 Cal (food calorie)
===
Energy Equivalents
These equivalents can be used in dimensional analysis to convert units
Heat Capacity
Specific Heat Capacity (Cp) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C
Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C
Heat Capacity
High Heat Capacity Low Heat Capacity
Takes a large amount of energy to noticeably change temp
Small amount of energy can noticeably change temperature
Heats up slowlyCools down slowlyMaintains temp better with small condition changes
Heats up quicklyCools down quicklyQuickly readjusts to new conditions
A pool takes a long time to warm up and remains fairly warm over night.The air warms quickly on a sunny day, but cools quickly at night
A cast-iron pan stays hot for a long time after removing from oven.Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly
What things affect temperature change?
Heat Capacity of substanceThe higher the heat capacity, the slower the
temperature changeMass of sample
The larger the mass, the more molecules there are to absorb energy, so the slower the temperature change
TCmH p
Energy added or removed
Mass of sample
Specific heat capacity of substance
Change in temperature
Positive & Negative T
Change in temperature (T) is always T2 – T1 (final temperature – initial temperature) If temperature increases, T will be positive
A substance goes from 15°C to 25°C. 25°C - 15°C = 10°CThis is an increase of 10°C
If temperature decreases, T will be negativeA substance goes from 50°C to 35°C35°C – 50°C = -15°CThis is a decrease of 15°C
Positive & Negative H
Energy must be put in for temperature to increaseA “+” T will have a “+” H
Energy must be removed for temperature to decreaseA “-” T will have a “-” H
Example
Example:If 285 J is added to 45 g of water at 25°C, what is the
final temperature? Cp water = 4.18 J/g°C
Example
TCmH p
H = change in energym = massCp = heat capacity T = change in temperature (T2 - T1)
T2 = 27°C
CTCgJgJ
2518.445285 2
225
18.445
285 TC
CgJg
J
Example:If 285 J is added to 45 g of water at 25°C, what is the
final temperature? Cp water = 4.18 J/g°C
CT
CgJg
J
2518.445
2852
Let’s Practice #1
Example:How many joules must be
removed from 25 g of water at 75°C to drop the
temperature to 30°? Cp water = 4.18 J/g°C
Let’s Practice #1
TCmH p
H = change in energym = massCp = heat capacity T = change in temperature (T2 - T1)
H = - 4703J
CCgJgH
0.750.3018.40.25
Example:How many joules must be
removed from 25.0 g of water at 75.0°C to drop the
temperature to 30.0°? Cp water = 4.18 J/g°C
Let’s Practice #2
Example:If the specific heat capacity of
aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g
sample at 15°C
Let’s Practice #2
TCmH p
H = change in energym = massCp = heat capacity T = change in temperature (T2 - T1)
T2 = 31.2°C
CTCgJgJ
0.15900.00.30437 2
20.15
900.00.30
437 TC
CgJg
J
Example:If the specific heat capacity of
aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g
sample at 15.0°C
CT
CgJg
J
0.15900.00.30
4372
Calorimetry
1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes
This is also referred to as the Law of Conservation of Energy
Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa
Calorimetry
Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system
Hsurroundings = - Hsystem
Because of the Law of Conservation of Energy,The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.
(m×Cp×T)surroundings = - (m×Cp×T)system
Don’t forget the “-” sign on one sideMake sure to keep all information about surroundings together and all information about system together—you can’t mix and match!
Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature
Two objects at different temperatures
So you know that T2 for the system is the same as T2 for the surroundings!
An example of CalorimetryExample:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of
the metal?
Metal:m = 23.8 gT1 = 100.0°CT2 = 32.5°CCp = ? Water:m = 50.0 gT1 = 24°CT2 = 32.5°CCp = 4.18 J/g°C Cp = 1.04 J/g°C
watermetal HH
waterpmetalp TCmTCm
CC
CgJgCCCg p
5.245.3218.40.500.1005.328.23
CCg
CCCg
JgC p
0.1005.328.23
5.245.3218.40.50
An example of CalorimetryExample:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of
the metal?
Let’s Practice #3Example:
A 10.0 g of aluminum (specific heat capacity is 0.900 J/g°C) at 95.0°C is placed in a
container of 100.0 g of water (specific heat capacity is 4.18 J/g°C) at 25.0°. What’s the
final temperature?
Example:A 10.0 g of aluminum (specific heat capacity
is 0.900 J/g°C) at 95.0°C is placed in a container of 100.0 g of water (specific heat capacity is 4.18 J/g°C) at 25.0°C. What’s
the final temperature?
Metal:m = 10.0 gT1 = 95.0°CT2 = ?Cp = 0.900 J/g°C Water:m = 100.0 gT1 = 25.0°CT2 = ?Cp = 4.18 J/g°C T2 = 26.5 °C
watermetal HH
waterpmetalp TCmTCm
CT
CgJgCTCg
Jg
0.2518.40.1000.95900.00.10 22
Let’s Practice #3
CTCT 0.25418950.9 22
104504188550.9 22 TT
113050.427 2 T
0.42711305
2 T
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