Lecture 24Heat Transfer

Lecture 23Internal Energy, Heat and phase

changes.

Internal Energy U

Internal energy is the total energy of all molecules in an object.

Ideal Gas: U = < KE

total> = N ½m <v 2>

U =32

NkT N: total number of molecules

k = 1.38×10−23 J (Boltzman constant)

U =32

nRT n: number of moles

R = 8.314 J/mol K (universal gas constant)

Heat

Energy is transferred spontanuously due to a temperature difference between two objects. There is no work involved here. This transferred energy is called heat.

Units:

SI: J (Joules)

cal (calorie) 1 cal = heat required to raise the temperature of 1 g of water from 14.5°C to 15.5°C

Specific heat

How much heat is needed to change by ΔT the temperature of a mass m of material X?

XQ mc T X specific heatc

water: c = 1 cal//(g °C) = 4186 J/(kg K)

iron: c = 470 J/(kg K)

cX does have some temperature dependence, but very small (ie, negligible for this course)

Definition of calorie!

Water has a very high specific heat.

It’s “hard” to increase the temperature of water.

Example: Kettle

Your electric kettle is labeled 2000 W. Estimate how long it will take to boil enough water to fill your 0.5 liter thermos if water comes out of the tap at 15°C.

Example: Kettle

Your electric kettle is labeled 2000 W. Estimate how long it will take to boil enough water to fill your 0.5 liter thermos if water comes out of the tap at 15°C.

Assuming that all the heat produced by the kettle is used to warm up the water,

water water waterQ m c T Heat needed to warm up the water:

Qwater = (0.5 Liters 1 kg1 Liter )(4186 J

kg∘C ) (100−15) ∘C = 1.78×105 J

t =QP

=1.78×105 J

2000 W= 89 s = 1.5 min

But of course in reality it will be a little longer. What are we neglecting?

Heat produced by the internal resistance of the kettle also warms up:

• air (negligible if kettle has a lid and is well insulated)

• kettle (at least internal wall)

water wall air producedQ Q Q P t

Energy released in resistor

Energy absorbed by water

Energy absorbed by wall

Energy absorbed by air

Flow of energy

What does specific heat depend on?

Temperature = average kinetic energy of particles

Degrees of freedom = ways to move, ie, to increase kinetic energy for the same temperature:

Example:

A molecule of helium is made of one atom. It can basically just “bounce around” in 3 directions (3 degrees of freedom.)

A molecule of hydrogen is made of two atoms. It can bounce around (3 dof) and it can also rotate (+2 dof, total 5 degrees of freedom)

Molar mass: Heavier molecules store require more additional energy to increase their average speed.

Phases or states of matter

Three basic states of matter:

•Solid

•Liquid

•Gas

A phase change involves a critical change in the microscopic structure of matter.Example:

Ice to water: Lattice disappears, molecules become free to move around.

Analysis: Phase changes for water

Q

T1 kg of ice is placed on a pan on the stove. Plot temperature and heat supplied by stove.

Ice

Ice + Liquid

Liquid Gas

Liquid + Gas

334 kJ(Latent) heat of fusion

2256 kJ

(Latent) heat of vaporization

100°CBoiling point

0°CMelting point

ACT: Specific heats

Q

T

Ice Liquid Gas

Which water phase has the largest specific heat?

A. Ice

B. Liquid water

C. Steam

ACT: Specific heats

Q

T

Ice Liquid Gas

Which water phase has the largest specific heat?

A. Ice

B. Liquid water

C. Steam

Small slope, large c

Q = mcΔ T ⇒ slope =1

mc

ice waterJ JIndeed: 2100 4186

kg K kg Kc c

Latent heat

Water: Lfusion = 334 kJ/kg Lvaporization = 2256 kJ/kg

This is the heat for 1 kg…

This energy is not used to increase the kinetic energy of the particles (does not increase the temperature) but to change the structure of matter.

During a phase change, two or more phases coexist in dynamic equilibrium. Examples: Vapor and liquid water exactly at 100°C

Ice and liquid water exactly at 0°C

Vapor/Liquid water/Ice at the triple point (273.16 K and 610 Pa)

In-class example: Ice melting

How much heat is needed to turn 10 g of ice at -5 °C into liquid water at 20 °C?

A. 105 J

B. 420 J

A. 837 J

B. 3330 J

C. 4272 J

In-class example: Ice melting

How much heat is needed to turn 10 g of ice at -5 °C into liquid water at 20 °C?

A. 105 J

B. 420 J

A. 837 J

B. 3330 J

C. 4272 J

total 4272 JQ

(mostly from melting)

Qtotal = Qwarm ice to 0∘C + Qmelt ice + Qwarm water to 20∘ C

Qwarm ice to 0 ∘C = (0.010 kg )(2100 Jkg∘C ) [0−(−5∘C ) ] = 105 J

Qmelt ice = (0.010 kg )(333kJkg ) = 3330 J

Qwarm water to 20 ∘C = (0.010 kg )(4186 Jkg∘C )(20 ∘C−0 ) = 837 J

Follow-up example: Iced coffee

10 g of ice at -5°C are added to 30 ml of hot coffee inside a thermos that is then tightly closed. After the system reaches equilibrium, the temperature of the mix is 20°C. What was the initial temperature of the coffee?

Because the system is thermally isolated (closed thermos), the hot coffee is the only source of energy, so it must provide the necessary 4272 J of heat.

4272 J

For the coffee, this is a decrease in energy

Qto cool coffee = mcoffeecwater (20 ∘C−T initial)

T initial = 20 ∘C−Qto cool coffee

mcoffeecwater

= 20 ∘C−−4272 J

(30×10−3 kg )(4186 Jkg ∘C )

= 54 ∘C

Calorimeter

System: Water + Inner vessel + substance (placed in water)

Environment: Every thing else;air space, outer vessel, etc.

is isolated (by very poor conducting air apace, Insulated lid and insulated ring) from the environment

Calorimeter is used to determine specific heat of substances

How does it work?1- Heat substance to a known T

s.

2- Place it in the water (at known lower temperature Tw ).

3- Stirr to insure equal distribution.4- Read thermometer when temperature is not changing anymore.5- Calculate c of substance from (Q

gained = Q

released).

10 g of ice at -5°C are added to 30 ml of hot coffee inside a thermos that is then tightly closed. After the system reaches equilibrium, the temperature of the mix is 20°C. What was the initial temperature of the coffee?

Energy balance:

+ + + −Energy is absorbed Energy is released

Energy released by coffee

Energy absorbed by ice (warming)

Energy absorbed by liquid water (warming)

Flow of energy: from hot object to cold object

Energy absorbed by ice (melting)

Qwarm ice to 0∘ C + Qmeltting ice + Qwarm water to 20∘ C + Qcool coffee = 0

Dealing with phase changes,the algorithm:

Let’s say you want to combine ice with something else that has T > 0 ºC and find the final temperature (Tf).

Step 1: Calculate Q1 needed to warm ice to 0 ºC: Q1 =micecice(0 – Tice) (positive)and Q4 needed cool substance to 0 ºC: Q4 =mothercother(0 –Tother) (negative)

if |Q4|<Q1 , then no ice melts and Tf < 0: micecice(Tf – Tice) = mothercother(Tother –Tfinal)

else: Step 2: Calculate Q2 needed to melt all ice: Q2 = miceLice-water (positive)

if Q1 <|Q4|<Q1+Q2, then only some ice melts (fraction of mice say mmelt) and Tf = 0:

micecice(0 –Tice) + mmeltLice-water = mothercother(Tother – 0)

one can solve for mmelt given all other parameters.

else ( i.e. |Q4|>Q1+Q2):

Step 3: Calculate Q3 needed to warm all melt ice to Tf (Tf > 0): Q3 = micecwater(Tf – 0) (positive)

micecice(0 –Tice) + miceLice-water + micecwater(Tf – 0) = mothercother(Tother –Tf)

Mechanisms of heat transfer

Conduction: Two surfaces in contact, or within a body.

Convection: Particles move between regions of different temperature and carry energy

Radiation: Energy carried by electromagnetic waves. Can happen through vacuum!

Conduction

Metal bar of length L and cross-sectional area A between two sources at THOT and TCOLD.

k = thermal conductivity

Qt

= kATH−TC

L

Also, the rate of transfer of energy decreases with length.

What transfers energy are the collisions between faster and slower molecules, and the number of collisions increases with cross-section area, A.

This formula “makes sense”:

Thermal conductivity is the efficiency of collisions to exchange energy

Thermal conductors and insulators

Conductors (high k) Metals 50-500 W/(m K)

Insulators (low k) Wood 0.12-0.04 W/(m K)Air (still) 0.024 W/(m K)Styrofoam 0.01 W/(m K)

Of course, there are no perfect thermal conductors or insulators.

How low can you go?????…. “Aerogel”: 8x10-5 W/(m K)What’s aerogel….

• a man-made substance formed by specially drying a wet silica gel, resulting in a solid mesh of microscopic strands.

• used on space missions to catch comet dust • the lightest material know to man, according to the Guiness Book of

World Records -- really, really light. It is ~98% porous, and yet it is quite rigid…

http://en.wikipedia.org/wiki/Aerogel

http://upload.wikimedia.org/wikipedia/commons/6/69/Aerogelflower_filtered.jpg

Aerogel

2.5 kg brick

2 g aerogelk = 8 x 10-5 W/m-K

DEMO: Water is also a poor

conductor

Example: Window

If it’s 22 °C inside and 0 °C outside, what is the heat flow through a glass window of area 0.3 m2 and thickness 0.5 cm?

The thermal conductivity of glass is about 1 W/m-K.

Example: Window

If it’s 22 °C inside and 0 °C outside, what is the heat flow through a glass window of area 0.3 m2 and thickness 0.5 cm?

The thermal conductivity of glass is about 1 W/m-K.

H = k AL (TH−TC ) = (1 W

mK )( 0.3 m2

5×10−3m )(22 K ) = 1320 W

Example: Double pane

How much heat is lost (per second) through a double-pane version of that window, with a 0.5-cm air gap? The thermal conductivity of air is about 0.03 W/(m K).

Example: Double pane

How much heat is lost (per second) through a double-pane version of that window, with a 0.5-cm air gap? The thermal conductivity of air is about 0.03 W/(m K).

We can pretty safely ignore the glass, which has a much higher conductivity than air. H is limited by conduction across the air gap.

Note: larger air gaps don’t always work better because convection currents swirl the hot and cold air around.

(<< 1320 W)

The heat current is limited by the air gap.

H = k AL (TH−TC ) = (0.03 W

mK )( 0.3 m2

5×10−3m )(22 K ) = 39.6 W

Convection

Particles move between regions of different temperature.

Mathematically… waaaay beyond 221

Example: Wind chill

Radiation

All pieces of matter emit electromagnetic radiation– typically long-wavelength, infrared– For hot bodies, radiation is in visible wavelengths e.g. light-bulbs

Stefan-Boltzmann Law

Rate of radiation of heat for a body a temperature T:

A : area of surface of bodye : emissivity of surface, empirical (e.g e = 0.3 for Cu): Stephan-Boltzmann constantT in Kelvins σ = 5.67×10−8 J

m2 s K4

Qt

= ϵσAT 4

Radiation is in all wavelengths. Depending on T, most heavily in one wavelength interval than other. This is NOT described by this law

DEMO: Lighting a

match

ACT: Radiating book

A book at room temperature in this room radiates energy. Yet the temperature of the book does not go down: why not?

A. Radiation stops when book is at room temperatureB. Heat is also absorbed by the book as radiation from the

roomC. There must be a heat-source in the book

ACT: Radiating book

A book at room temperature in this room radiates energy. Yet the temperature of the book does not go down: why not?

A. Radiation stops when book is at room temperatureB. Heat is also absorbed by the book as radiation from the

roomC. There must be a heat-source in the book

if T room = Tbook ⇒Qradiated

t+

Qabsorbed

t=0 ⇒ temperature unchanged

Qradiated

t= Abook ϵbook σT book

4

Qabsorbed

t= Abook ϵbook σT room

4

Emission and absortion

The term radiation refers to emission and absorption of energy:

Emissivity:

Є = 1 black body: ideal emitter and ideal absorber (absorbs all the radiation that strikes it)eg. Sun

Є = 0 ideal reflector (absorbs no radiation and does notemit either)eg. Mirrored lining inside thermos bottles

Qnet

t= Aϵσ (T 4

− T environment4 ) Radiation:

Qnet

t>0

Absorbtion: Qnet

t<0

In-class example: Sitting inside a fridge

You are sitting in your shorts inside a cold (0 °C) room. At what rate are you transferring heat to the room by radiation? (Take the worst case scenario with e = 1 and A = 2 m2)

A. 0

B. 2.410-4 W

C. 404 W

D. 8.9 104 W

E. None of the above

In-class example: Sitting inside a fridge

You are sitting in your shorts inside a cold (0 °C) room. At what rate are you transferring heat to the room by radiation? (Take the worst case scenario with e = 1 and A = 2 m2)

A. 0

B. 2.410-4 W

C. 404 W

D. 8.9 104 W

E. None of the above

Flow from body to room

Note: In just 10 s, this is a loss of 4000 J ~ 1000 cal = 1 Cal (food calorie.) We burn food to stay warm!

= (2 m2)(5.67×10−8 J m2 s K4 ) (3094

−2734 )K4

= 404 W

Qnet

t= Aϵσ (T 4

− T environment4 )

Radiation: Effective Area

If you are in the sunlight, the Sun’s radiation will warm you. In general, you will not be perfectly perpendicular to the Sun’s rays, and will absorb energy at the rate:

Q

t= ϵ σ Aeffective (T sun

4−Tbody

4 )

Aeffective = Abody cos θ

Radiation: Effective Area

This cosθ effect is also responsible for the seasons.