Part One Heat and Temperature Part One Heat and Temperature

Part OneHeat and

Temperature

Part OneHeat and

Temperature

What is temperature?

What does it mean to have a temperature of 0 C?

Is temperature the same

thing as heat?

Temperature is a measure of how “hot” or “cold” something is.

Temperature is measured in arbitrary units, like Fahrenheit or Celsius.

Temperature is proportional to the average kinetic energy of the molecules of the substance.

Heat is the thermal energy transferred from a hot object to a cold object.

Heat is measured in energy units -- Joules or calories.

Heat has the symbol q and is calculated using …

q = mcT

q = mcT

Quantity of heat

mass

specific heat capacity

temperature change

q = mcT

Quantity of heat

specific heat capacity

The specific heat capacity of water

is 4.18 J/gC

q = m c T

How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C?

q = (25.6g)(4.18J/gC)(30.0C)

q = 3210 J

q = m c T

What is the final temperature of 27.0 grams of liquid water, initially at 0C, after it absorbs 700.0 J of energy?

Hint: start by solving for T.

=qT m c

Answer: 6.20 C

Part TwoPhase Changes

Part TwoPhase Changes

We now know that heat is either absorbed or released during a phase change.

A process that gives off heat is called

exothermic.

A process that absorbs heat is called endothermic.

Exothermic:

Endothermic:

FreezingCondensationDeposition

Melting (fusion)

VaporizationSublimationHeat is absorbed.

Heat is released.

Ice And melts.

Heat is absorbed by the ice.

Heat is absorbed by the ice.

… making liquid water

One gram of ice at 0C absorbs 334 J as it melts to form water at 0C.

water

Heat is released by the water as it freezes.

334 joules is released when one gram of water freezes at 0C.

Ice

Ice

Ice absorbs 334 J per gram as it melts at 0C

Water releases 334 J per gram as it freezes at 0C

Hotplate

Water absorbs

2260 J/g as it boils at 100 C

Steam releases 2260 J/g as it condenses at 100 C

The heat gained or lost in phase changes can be calculated using …

q = mHf q = mHv

Heat of fusion

(melting)

Heat of vaporization

The values for water are …

Hf = 334 J/g

Hv=2260 J/g

Heat of fusion

(melting)

Heat of vaporization

How much heat is absorbed by 150.0 g of ice as it melts at 0C?q = m Hf q = (150.0 g)(334 J/g)

q = 50,100 J or 50.1 kJ

How much heat is released by 20.0 grams of steam as it condenses at 100C?q = m Hv q = (20.0 g)(2260 J/g)

q = 45,200 J or 45.2 kJ

Part ThreePhase Diagrams

Part ThreePhase Diagrams

Temperature

Pre

ssur

e

The phase diagram has three distinct regions.

2

31

Hint: What happens to ice as temperature increases?

Which phase is in each region?

The point where all three phases exist in equilibrium is called the

Temperature

S L

G

triple point. triple point.

Pre

ssur

e

At a pressure of 1 atm, most substances go through all three phases, as the temperature increases,

Temp.

S L

G1 atm

Solids melt to form liquids, which vaporize to form gases.

Temp.

S L

G1 atm

MP BP

Notice the melting point and boiling point.

But the phase diagram for CO2 is a little different.

Temperature

S L

G1 atm

Notice that the triple point is above 1 atm.

5 atm

At 1 atm CO2 goes directly from solid to vapor as the temperature increases.

Temperature

S L

G1 atm

The sublimation point is –78.5 C

What phase change is occurring?

Melting (fusion)

Temperature

Pre

ssur

e

S L

G

What phase change is occurring?

Temperature

Pre

ssur

e

VaporizationS L

G

What phase change is occurring?

Temperature

Pre

ssur

e

CondensationS L

G

What phase change is occurring?

Temperature

Pre

ssur

e

SublimationS L

G

What phase change is occurring?

Temperature

Pre

ssur

e

Liquefying a gas by increasing the pressure.

S L

G

Part FourHeating and Cooling

Curves

Part FourHeating and Cooling

Curves

Look at the different regions of the heating curve for water.

0

100

Time

Temp

Ice

Ice andwater

Water

Steam

Phase changes?

Water andsteam

Part FiveCalorimetry and

Specific Heat Capacity

Part FiveCalorimetry and

Specific Heat Capacity

Calorimetry is a collection of laboratory procedures used to investigate the transfer of heat.

In calorimetry experiments, one might be looking for a final temperature or a specific heat capacity.

What is the law of conservation of energy?

Energy is neither created nor destroyed, only changed in form.

The law of conservation of energy suggests that the heat lost by the hot object as it cools is equal to the heat gained by the cool water as it warms up.

Investigate:

To put it mathematically:

qlost = -qgained

And since q = mcT then

mocTo = -mwcTw

Heat lost by the hot object =

Heat gained by the cold water

Investigate:The convention for T is final temperature minus initial temperature or Tfinal – Tinitial

moc(Tf -Ti) = -mwc(Tf -Ti)Use your algebra skills, to solve for Tf , the final temperature.

mhcTo = -mccTw

becomes

Specific heat capacity …

• …varies from one substance to another.

• …a measure of how much heat something can “hold”.

• …the amount of heat needed to raise one gram of a substance by one Celsius degree.

Specific heat capacity lab suggestions:

1. Heat a metal to a known temp. 2. Transfer the metal to a known

quantity of water at a known temperature.

3. Measure the equilibrium temperature.

4. Use qlost = -qgained to compute the specific heat of the metal.

hotplate

Get the initial temperature of the metal.

The temperature of boiling water.

metal

Get initial temp of water in calorimeter cup.

Transfer the metal to the calorimeter.

Continue stirring.

Data: Mass of metalInitial temp of metal

Mass of waterInitial temp of water

Final temp of water and metal

qlost = -qgained

mmcmTm = -mwcwTw

-mwcwTw

mmTm

cm =

Mass of metal 40.0 gInitial T of metal 98.0 CMass of water in calorimeter 60.0 gInitial T of water 20.0 CFinal T of water and metal 22.9 C

Calculate the specific heat capacity of the

metal.

Aluminum 0.900Bismuth 0.123Copper 0.386Brass 0.380Gold 0.126Lead 0.128Silver 0.233Tin 0.225Zinc 0.387Mercury 0.140Ethanol 2.400Water 4.186Ice 2.050

Substancec in

J/g KTable of selected specific heats.

What is the unknown

metal?