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 slowly

Cools down slowly

Maintains temp better with small condition changes

Heats up quickly

Cools down quickly

Quickly 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 change

Mass of sampleThe 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 DT

Change in temperature (DT) is always T2 – T1 (final temperature – initial temperature) If temperature increases, DT 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, DT will be negativeA substance goes from 50°C to 35°C35°C – 50°C = -15°CThis is a decrease of 15°C

Positive & Negative DH

Energy must be put in for temperature to increaseA “+” DT will have a “+” DH

Energy must be removed for temperature to decreaseA “-” DT will have a “-” DH

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

DH = change in energym = massCp = heat capacity DT = change in temperature (T2 - T1)

T2 = 27°C

CTCg

JgJ 2518.445285 2

225

18.445

285TC

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

DH = change in energym = massCp = heat capacity DT = change in temperature (T2 - T1)

DH = - 4703J

CCg

JgH 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

DH = change in energym = massCp = heat capacity DT = change in temperature (T2 - T1)

T2 = 31.2°C

CTCg

JgJ 0.15900.00.30437 2

20.15

900.00.30

437TC

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

DHsurroundings = - DHsystem

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×DT)surroundings = - (m×Cp×DT)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

0.427

113052 T