msu physics 231 fall 2015 1 physics 231 topic 13: heat alex brown dec 1, 2015

MSU Physics 231 Fall 2015 1

Physics 231Topic 13: Heat

Alex BrownDec 1, 2015


9th 10 pm attitude survey (1% for participation)

11th 10 pm last homework set

8th 10 pm correction for 3rd exam

10th 10 pm concept test timed (50 min)) (1% for performance)

17th 8-10 pm final (Thursday) VMC E100


According to the Ideal Gas Law, when the temperature is reduced at

constant pressure, the volume is reduced as well. The volume of the

balloon therefore decreases.

Clicker Quiz!

nRTPV =

a) it increases

b) it does not

change

c) it decreases

What happens to the volume

of a balloon if you put it in the

freezer?


Key Concepts: Heat

Heat and Thermal Energy

Heat & its units

Heat Capacity & Specific HeatThermal Equilibrium

Equipartition theorem

Phase ChangesLatent heat of fusion, vaporization

Conduction, Convection, and Radiation

Thermal Conductivity

Stefan-Boltzman law

Covers chapter 13 in Rex & Wolfson


Thermal equilibrium

Low temperatureLow kinetic energyParticles move slowly

High temperatureHigh kinetic energyParticles move fast

Thermal contact

Transfer of kinetic energy

Thermal equilibrium: temperature is the same everywhere


Heat

Heat: The transfer of thermal energy between objectsbecause their temperatures are different.

Heat: energy transfer Symbol: Q

Units: Calorie (cal) or Joule (J) 1 cal = 4.186 J (energy needed to raise

1g of water by 10C)


Heat transfer to an object

Q = c m T

Energy transfer(J or cal)

Specific heatJ/(kgoC) or cal/(goC)

Mass of object

Change in temperature

The amount of energy transfer Q to an object with mass m when its temperature is raised by T:


ExampleA 1 kg block of Copper israised in temperature by10oC. What was the heattransfer Q?

Answer:Q = c m T = (387)(1)(10) = 3870 J

Q = (0.092)(1000)(10) = 924.5 cal

1 cal = 4.186 J

Substance Specific Heat

J/kg oC

Specific Heat

cal/g oC

aluminum 900 0.215

copper 387 0.092

water 4186 1.00


ProblemA block of Copper is dropped from a height of 10 m. Assuming that all the potential energy is transferred into internal energy when it hits the ground, what is the raise in temperature of the block? ccopper=387 J/(kgoC) Potential energy: mgh (Joules)

All transferred into heat Q = cm T

mgh = cm T

T = gh/c = (10)(9.81)/387 = 0.25oC


CalorimetryIf we connect two objects with different temperatureenergy will transferred from the hotter to the coolerone until their temperatures are the same. If the system is isolated:

Energy flow into cold part = Energy flow out of hot part

mc cc ( Tf - Tc) = mh ch (Th - Tf)

the final temperature is: Tf =

note that T can be in Kelvin or Centigrade

mc cc Tc + mh ch Th

mc cc + mh ch

Tc (cold) Th (hot)

Tf (final)


Clicker Question

A block of iron that has been heated to 1000C is droppedin a glass of water at room temperature (200C). After the temperatures in the block and the water have become equal:

a) The water has changed more in temperature than the iron block.b) The water has changed less in temperature than the iron blockc) the temperatures of both have changed equallyd) I need more info to say anything!


Heating Water with a Ball of Lead

A ball of Lead at T=100oC with mass 400 g is dropped in aglass of water (0.3 L) at T=200C. What is the final temperature of the system?

cwater=1 cal/g oC clead=0.03 cal/g oC water=103kg/m3

Tfinal =

= [(300) (1) (20) + (400)(0.03)(100)] / [(300)(1) + (400) (0.03)]

= 7200/312 = 23.1 oC

mwatercwaterTwater + mleadcleadTlead

mwatercwater + mleadclead


And anotherA block of unknown substance with a mass of 8 kg, initiallyat T=280K is thermally connect to a block of copper (5 kg)that is at T=320 K (ccopper=0.093 cal/g0C). After the system has reached thermal equilibrium the temperature T equals 290K. What is the specific heat of the unknown materialin cal/goC?

????

copper

Cunknown = mcopperccopper(Tcopper-Tfinal ) munknown (Tfinal -Tunknown)

Cunkown = (5000) (0.093) (320-290) = 0.17 cal/goC 8000 (290-280)


Mixing 3 liquidsThree different liquids are mixed together in a calorimeter. The masses, specific heats and initial temperatures of the liquids are: m1 = 475 g,

m2 = 355 g,

m3 = 795 g.

What will be the temperature of the mixture in Celsius?

c1 = 225 J/kgC,

c2 = 500 J/kgC,

c3 = 840 J/kgC.

T1 = 24.5 °C,

T2 = 53.5 °C,

T3 = 81.5 °C.

Tf = =c1m1T1+c2m2T2+c3m3T3

c1m1+c2m2+c3m3

225x475x24.5 + 500x355x53.5 + 840x795x81.5

225x475 + 500x355 + 840x795= 69.9oC

Like with two substances, the final temperature is a weightedaverage of T1,T2 and T3 with the c’s and m’s being the weights


Internal EnergyIn chapter 12: The internal (total) energy for an idealgas is the total kinetic energy of the atoms/particlesin a gas.

For a non-ideal gas: the internal energy is due to kineticand potential energy associated with theinter-molecular potential energy

PErPE: negative!


phase changes

gas (high T)

liquid (medium T)

solid (low T)Q=cgasmT

Q=cliquidmT

Q=csolidmT

Gas liquid

liquid solid


phase changes

SOLID to Gas and GAS to LIQUID

DURING THESE PHASE TRANSITIONS THE TEMPERATURE DOES NOT CHANGE

AND SO THE KINETIC ENERGY DOES NOT CHANGE .

ALL ADDED HEAT GOES TO CHANGING PE“breaking the chemical bonds”


phase changes

Gas liquid

When heat is added to a liquid, potential energy goes to zero - the energy stored in the stickiness of the liquid is taken away.

When heat is taken from a gas, potential energy goes into the stickiness of the fluid

liquid solid

When heat is added to a solid to make a liquid, potential energy in the bonds between the atoms become less.

When heat is taken from a liquid, the bonds between atoms becomes stronger - potential energy is more negative.


Okay, the Temperature does not change in a phase transition!

But what is the amount of heat added to make the phase transition?

Gas liquid

Qgas-liquid = m Lv

Lv=latent heat of vaporization (J/kg or cal/g)depends on material.(energy required to vaporize)

Gas to liquid Q flows outLiquid to gas Q flows in

m = mass


solid liquid

Qliquid-solid = m Lf

Lf=latent heat of fusion (J/kg or cal/g)depends on material (energy required to liquify)

Liquid to solid Q flows outSolid to liquid Q flows in

m = mass


phase changes

gas (high T)

liquid (medium T)

solid (low T)Q=cgasmT

Q=cliquidmT

Q=csolidmT

Gas liquid

liquid solid

Q=m Lv Q=m Lf


Clicker Question!

Ice is heated steadily and becomes liquid and then vapor.During this process:

a) the temperature rises continuously.b) when the ice turns into water, the temperature drops for a brief moment.c) the temperature is constant during the phase transformationsd) the temperature cannot exceed 100oC


T (oC)

0

100

ice ice+

wate

r wate

r

wate

r+st

eam

steam

THE PHASE TRANSFORMATIONS OF WATER


Ice with T=-30oC is heatedto steam of T=1500C.How much heat (in cal) hasbeen added in total?cice=0.5 cal/goCcwater=1.0 cal/goCcsteam=0.480 cal/goCLf=540 cal/gLv=79.7 cal/gm=1 kg=1000g

A) Ice from -30 to 0oC Q=1000*0.5*30= 15000 calB) Ice to water Q=1000*540= 540000 calC) water from 0oC to 100oC Q=1000*1.0*100=100000 calD) water to steam Q=1000*79.7= 79700 calE) steam from 100oC to 1500C Q=1000*0.48*50=24000 cal TOTAL Q= =758700 cal

T (oC)

0

100ic

e

ice+

wate

r

wate

r

wate

r+st

eam

steam


Question

Given: Lf=6.44x104 J/kg Tmelt=1063oC cspecific=129 J/kg0C

A block of gold (room temperature 200C)is found to just melt completely after supplying 4x103 J of heat. What was the mass of the gold block?

Properties of gold c = 129 J/(kg oC), melting point = 1063 oCLf = 6.45x104 J/kg

Q = c m T + m Lf

= (129)(m)(1063-20) + (m)(6.45x104) = 2.0x105 m4000 = 2.0x105 mm = 0.02 kg


How can heat be transferred?


Conduction

Touching different materials: Some feel cold, othersfeel warm, but all are at the same temperature…


Thermal conductivitymetal wood

T=370CT=370C

T=200C

T=200C

The heat transferin the metal is much faster thanin the wood:(thermal conductivity)


Heat transfer via conduction

Conduction occurs if there is a temperature difference betweentwo parts of a conducting medium

Rate of energy transfer P

P = Q/t (unit Watt = J/s)

P = k A (Th-Tc)/x = k A T/x

k: thermal conductivity Unit: J/(m s oC)

Metals k ~ 300 J/(m s oC)Gases k ~ 0.1 J/(m s oC)Nonmetals k ~ 1 J/(m s oC)

Th Tc

x

A


ExampleA glass window (A=4m2, x=0.5cm) separates a living room (T=20 oC) from the outside (T=0 oC).

A) What is the rate of heat transfer throughthe window?, kglass=0.84 J/(m s oC)

B) By what fraction does it changeif the surface becomes 2x smallerand the outside temperature drops to -20 oC?

A) P = k A T/x = (0.84)(4)(20)/0.005 = 13440 Watt

B) Porig= k A T/x Pnew= k(0.5A)(2T)/x = Porig

The heat transfer is the same

Th Tc

A

x


Another one

An insulated gold wire (i.e. no heat lost to the air) is atone end connected to a heat reservoir (T=1000C) and at theother end connected to a heat sink (T=200C). If its lengthis 1m and P=200 W what is its cross section (A)?kgold = 314 J/(m s oC).

P = k A T/x = (314)(A)(80)/1 = 25120 A = 200

A = 8.0x10-3 m2


Clicker Quiz!Given your experience of what

feels colder when you walk on it,

which of the surfaces would

have the highest thermal

conductivity?

a) a rug

b) a steel surface

c) a concrete floor

d) has nothing to do with thermal conductivity

The heat flow rate is k A (T1 − T2)/x. All things being

equal, bigger k leads to bigger heat loss.

From the book: Steel = 40, Concrete = 0.84,

Human tissue = 0.2, Wool = 0.04, in units of J/(s.m.C°).


Multiple Layers

iii

ch

kL

TTA

t

QP

)/(

)(

Th Tc

A

L1 L2 (x)

k1 k2


Insulation

L1L2 L3

insi

de

ThTc

iii

ch

kL

TTA

t

QP

)/(

)(

A house is built with 10cm thick wooden walls and roofs.The owner decides to install insulation. After installationthe walls and roof are 4cm wood + 2cm insulation + 4cm wood.If kwood=0.10 J/(ms0C) and kinsulation=0.02 J/(ms0C), by whatfactor does he reduce his heating bill? Pbefore = A T/[0.10/0.10] = A TPafter = A T/[(0.04/0.10) + (0.02/0.02) + (0.04/0.10)] = A T/1.8 Almost a factor of 2 (1.81) !


Convection

T high low


Radiation (photons)

Nearly all objects emit energy through radiation:

energy radiated per second

P = A e T4 : Stefan’s law

= 5.6696x10-8 W/m2K4

A: surface areae: object dependent constant emissivity (0-1)T: temperature (K)


Emissivity

Ideal reflectore=0no energy is absorbed

Ideal absorber (black body)e=1all energy is absorbedalso ideal radiator!


A BarbecueThe coals in a BBQ cover an area of 0.25m2. If the emissivity of the burning coal is 0.95 and their temperature 5000C, how much energy is radiated everyminute?

P = A e T4 (J/s) = 5.67x10-8 * 0.25 * 0.95 * (773)4 = 4808 (J/s)

1 minute: Q = 2.9x105 J (enough to boil off one liter of water)


Net Power Radiated

If an object would only emit radiation it would eventuallyhave 0 K temperature. In reality, an object emits ANDreceives radiation.

PNET = Ae (T4-T04)

where

T: temperature of objectT0: temperature of surroundings.


ExampleThe temperature of the human body is 370C. If theroom temperature is 200C, how much heat is givenoff by the human body to the room in one minute?Assume that the emissivity of the human body is 0.9and the surface area is 2 m2.

P = A e (T4-T04)

= 5.67x10-8 * 2 * 0.9 * (310.54 - 293.54) = = 185 J/s

Q = P * t = 185 * 60 = 1.1x104 J


Black bodyA black body is an object that absorbs all electromagnetic radiation that falls onto it. They emit radiation, depending on their temperature. If T<700 K, almost no visible light is produced (hence a ‘black’ body).

The energy emitted from a black body: P=T4 with =5.67x10-8 W/m2K4

b=2.90×10−3 m K

Wien’s displacement constant


Infrared RadiationThe human body emits radiation in the infrared.

With a body temperature of T=(273+37 K) = 310 K the wavelength (lmax) of the thermal emission is 9.4 x10-5 m


An ExampleThe contents of a can of soda (0.33 kg) which is cooled to 4 oC is poured into a glass (0.1 kg) that is atroom temperature (200C). What will the temperatureof the filled glass be after it has reached full equilibrium(glass and liquid have the same temperature)?

cwater=4186 J/(kgoC) and cglass=837 J/(kg0C)

Tfinal=

= (0.33*4186*4 + 0.1*837*20) / (0.33*4186 + 0.1*837) = 4.9oC

mwatercwaterTwater + mglasscglassTglass

mwatercwater + mglasscglass


And anotherWater 0.5L1000C

1500C

A = 0.03m2 thickness = 0.5cm.

A student working for his exam feels hungry and starts boilingwater (0.5L) for some noodles. He leaves the kitchen when the water just boils. The stove’s temperature is 1500C. The pan’s bottom has dimensions given above. Working hard on the exam, he only comes back after half an hour. Is there still water in the pan? (Lv=540 cal/g, kpan=1 cal/(m s 0C)

To boil away m = 500g of water: Q = Lv(500)=270000 calHeat added by the stove:

P = kA T/x = (1)(0.03)(50)/0.005 = 300 cal/s

P=Q/t t = Q/P = 270000/300 = 900 s (15 minutes)He’ll be hungry for a bit longer…

msu physics 231 fall 2015 1 physics 231 topic 13: heat alex brown dec 1, 2015

Documents

oc msu physics

b physics

everywheremsu physics

cmsu physics

rex wolfson msu physics

final temperature

j energy

energy transfer q