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 (C) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C

Cw (C 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 T

Change in temperature (T) is always 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

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°? Cw = 4.18 J/g°C

Let’s Practice #1

€

H = m ×Cw × ΔT

H = change in energym = massCw = heat capacity of water T = change in temperature (Tf - Ti)

H = - 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

If the specific heat capacity of aluminum is 0.900 J/g°C, how much heat must be added to a

30.0 g sample at 15°C to increase its temperature to

30°C ?

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

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.

Calorimetry

Hsurroundings = - Hsystem

(m×C×T)surroundings = - (m×C×T)system

Calorimetry also uses the principle of thermal equilibrium.

Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature

Calorimetry

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 specific capacity of

the metal?

Metal:m = 23.8 gTi = 100.0°CTf = 32.5°CCm = ? Water:m = 50.0 gTi = 24°CTf = 32.5°CCw = 4.184 J/g°C

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?

Step 1: Find the Heat (ΔH) of the waterΔH = mCwΔT

ΔH = (50)(4.184)(32.5 – 24) ΔH = 1778.2 JStep 2: Remember that the water is the “surroundings”. So, the heat gained by the water is the same as the heat lost by the metal but opposite in sign!!

Step 3: Since you know ΔH of the metal system (-1778.2 J), use this the find Cm.

ΔH = mCmΔT

Now use the values for the metal!

-1778.2 = (23.8)Cm(32.5 – 100)-1778.2 = (23.8)(-67.5)Cm

(23.8)(-67.5)Cm = 1.11 J/goC

Let’s PracticeExample:

A 10.0 g of aluminum at 95.0°C is placed in a container of 100.0 g of water (C = 4.184

J/g°C) at 25.0°. The metal and water reach a final temperature of 26.5 oC. What is the

specific heat of the metal?