Ch. 11 – Heat

11.1 Heat and Internal Energy

Internal Energy U – the energy associated with the microscopic kinetic (translational,vibrational, and rotational) and potential (intermolecular forces) energies of the system.

Heat is a mechanism by which energy is transferred between a system and itsenvironment because of a temperature difference between them. We will thus not talkabout how hot something is, but how much energy Q is transferred to a system due to atemperature difference.

Heat is thermal energy, the ability to change the T of a system in contact.

Historical Unit: calorie

1 calorie is the energy necessary to raise the temperature of 1 g of water from14.5°C to 15.5°C.

Note: the familiar “Calorie” used in determining the chemical energy of food is actually 1kcal (1000 cal).

Mechanical Equivalent of Heat – 1 cal ! 4.186 J or 1 Cal ! 4186 J

Example: Problem 7

A 75.0 kg weight-watcher wishes to climb a mountain to work off the equivalent of alarge piece of chocolate cake rated at 500 (food) Calories. How high must the personclimb? (1 food Calorie = 103 calories).

11.2 Specific Heat

Some substances require more/less than 4.186 J to raise 1 g by 1°C (or 4186 J for 1 kg by1°C). We use specific heat to describe and quantify this fact.

specific heat c !

Q

m"TJ kg#1 !C#1

where !T = T

final" T

initial . NOTE ORDER!

By definition, the specific heat for water is c = 4186 J kg-1 °C-1.

So the energy required to raise the temperature of a system of mass m by "T is:

Q = mc!T

Sign Convention:

When T in a system increases, energy flowing into the system - Q and "T are positiveWhen T in a system decreases, energy flowing out of the system - Q and "T are negative

Application: Sea breezes

Specific heat of sand is lower than that of water. Daytime heating causes breeze from seato land as hot air rises over the sand.

Exercise: Problem #2

A 50 g sample of copper is at 25°C. If 1,200 J of energy is added to the copper by heat,what is its final temperature?

Quick Quiz 11.1Imagine you have 1 kg each of iron, glass, and water, and that all of the samples are at10°C. (a) Rank the samples from lowest to highest temperature after 100 J is added toeach by heat. (b) Rank them from least to greatest amount of energy transferred by heat ifenough energy is transferred so that each increase in temperature by 20°C.

11.3 CalorimetrySuppose we put a warm object into cooler water:

For any isolated system exchanging heatamongst its components:

Qcold= !Q

hotor Q

cold+ Q

hot= 0

Here,

Qcold

= Qwater

= mwc

wT ! T

w( )Q

hot= Q

substance x= m

xc

xT ! T

x( )m

wc

wT ! T

w( ) = !mxc

xT ! T

x( )so c

x=

mwc

wT ! T

w( )m

xT

x! T( )

Sign Convention:

The energy transfer Qhot is negative because energy is leaving the hot substance. The

minus sign in Qcold= !Q

hotinsures that Qcold

is positive, since it is the cold object that isgaining energy. Note that Qcold

+ Qhot

= 0 . This is may be easier to remember.

Example: Problem 15

An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27°C. A 400 gsample of silver at an initial temperature of 87°C is placed in the water. The stirrer is usedto stir the mixture gently until it reaches its final equilibrium temperature of 32°C.Calculate the mass of the aluminum cup.

Example: Problem 18

A combination of 0.250 kg of water at 20.0°C, 0.400 kg of aluminum at 26.0°C , and0.100 kg of copper that is at 100°C is mixed in an insulated container and allowed tocome to thermal equilibrium. Neglect any energy transfer to or from the container anddetermine the final temperature of the mixture.

10.4 Latent Heat & Phase Change

Phase Changes: vapor#liquid boiling/condensingliquid#solid melting/freezingvapor#solid sublimationsolid1#solid2

Energy associates with phase change Q = ±mL , where L is the latent heat of thesubstance for that particular phase change.

Sign Convention:

Q = +mL adding energy example : melting ice cube( )Q = !mL removing energy example : freezing water( )Heat of fusion – melting/freezingHeat of vaporization – boiling/condensingHeat of sublimation

Remember, in solving problems:

1. consistent units2. Q = mc!T only if no phase change

Q = ±mL

fusion (solid#liquid);

Q = ±mL

vapor (liquid#vapor)

3. remember ± signs, and that !T = T

final" T

initial

Example: Taking ice at T=-30.0°C to vapor at T=120.0°C (at constant pressure)

Part A. Warming ice, T changes Q = mcice!T

Part B. Melting ice, T constant Q = mL

fusion

Part C. Warming water, T changes Q = mcwater

!T

Part D. Vaporizing water, T constant Q = mL

vapor

Part E. Warming vapor. T changes Q = mcsteam

!T

Exercise: Problem 23

What mass of steam that is initially at 120°C is needed to warm 350 g of water and its300-g aluminum container from 20°C to 50°C?

11.5 Thermal Conduction

Conduction – net transfer of energy due to vibrational motions of atoms, molecules, and(in the case of electrical conductors) mobile electrons without a net flow of the material.

!=Q

"t= k

thermalconductivity

! A"T

"x= kA

Thotter

# Tcolder

"xhotter#cooler

$

%&'

()

!=Q

"t=

A Thotter

# Tcolder( )

Li

ki

Rii$!

i$

=A T

hotter# T

colder( )R

net

Can you think of another example where the flow goes inversely with the “resistance”?

Example: Problem #39

A copper rod and an aluminum rod of equal diameter are joined end to end in goodthermal contact. The temperature of the free end of the copper rod is held constant at100°C, and that of the far end of the aluminum rod is held at 0°C. If the copper rod is0.15m long, what must be the length of the aluminum rod so that the temperature at thejunction is 50°C?

11.6 Convection

Convection – heat transfer by macroscopic movement of matter.

Natural Convection – lower density creates buoyancy - heated roadways, campfires, etc.

Forced Convection – using fans, pumps, etc. – some heating systems

Mathematically, convection is very difficult to deal with – far beyond the scope of thiscourse! (Notice that the textbook has NO problems for this section!)

11.7 Radiation

All objects with T > 0 K radiate electromagnetic radiation at a rate of:

!= " AeT 4 Watts Stefan 's Law

" = 5.699x10#8 W m#2K #4

A = surface area

e = emissivityAn object in radiative “contact” with surroundings at temperature T0 will absorb anamount of energy

!abs= " AeT

04

So the net gain/loss of energy by radiation is

!

net= " Ae T 4 # T

04( )

(Actually it’s a little more complicated than this, because here we have assumed that “e”is the same for the absorption and emission – i.e. that the absorption and emission occurover the same wavelength ranges.)

Thermogram of a human.

Example: Problem 11.44Calculate the temperature at which a tungsten filament that has an emissivity of 0.25 anda surface area of 2.5x10-5 m2 will radiate energy at a rate of 25 W in a room where thetemperature is 22°C.

11.8 Resisting Energy Transfer

Thermos bottles (more appropriately calledDewar flasks) help thermally isolate theircontents thermally from the rest of theworld.

The Hubble Space Telescope reduces Tfluctuations from entering & exiting theEarth’s shadow using a blanket of highlyreflective insulation.

11.9 Global Warming & Greenhouse Gases

The low “e” in the infrared due to various gases raises the T of the earth above what itwould be in the absence of an atmosphere. This “greenhouse effect” is a misnomer, asreal greenhouses get most of their T increase from inhibiting convective energy transport.

CO2, one of the major greenhouse gases, ison the rise. A doubling of CO2 could leadto a global increase in T of 2°C (accordingto one estimate), which could have asignificant impact on the climate of theEarth.

In addition to CO2, CH4, N2O, and SO2 contribute.

2/3 of the greenhouse effect is due to H2O!! This was not included in climate modelsuntil about 10 years ago.

There is some evidence that the Earth is warming slightly in the past century (althoughnot all parts of the atmosphere seem to be doing the same thing, so it is complex).Changes in solar irradiation may also be contributing to this trend.