Lecture 33

General Physics (PHY 2130)

http://www.physics.wayne.edu/~apetrov/PHY2130/

•  Heat   Ideal gases   Phase transitions. Latent heat.

Lightning Review

Last lecture: 1.  Thermal physics

  Heat. Specific heat.

Review Problem: Doppler ultrasound is used to measure the speed of blood flow. If the speed of the red blood cells is v, the speed of sound in blood is u, the ultrasound source emits waves of frequency f, and we assume that the blood cells are moving directly toward the ultrasound source, then the frequency fr of reflected waves detected by the apparatus is given by the formula below. The reflected sound interferes with the emitted sound, producing beats. If the speed of red blood cells is 0.11 m/s, the ultrasound frequency used is 4.6 MHz, and the speed of sound in blood is 1570 m/s, what is the beat frequency?

3

The beat frequency is 21 fff −=Δ

Two waves of different frequency

Superposition of the above waves

4

Doppler ultrasound is used to measure the speed of blood flow. If the speed of the red blood cells is v, the speed of sound in blood is u, the ultrasound source emits waves of frequency f, and we assume that the blood cells are moving directly toward the ultrasound source, then the frequency fr of reflected waves detected by the apparatus is given by the formula below. The reflected sound interferes with the emitted sound, producing beats. If the speed of red blood cells is 0.11 m/s, the ultrasound frequency used is 4.6 MHz, and the speed of sound in blood is 1570 m/s, what is the beat frequency?

The beat frequency is given by the difference of frequencies of emitted and reflected ultrasound waves:

Specific Heat • Every substance requires a unique amount of energy per unit mass to change the temperature of that substance by 1° C •  directly proportional to mass (thus, per unit mass)

• The specific heat, c, of a substance is a measure of this amount

c = Cm=

Qm ΔT

Units

SI Joule/kg °C (J/kg °C)

CGS Calorie/g °C (cal/g °C )

Notes: Heat and Specific Heat

• Q = m c ΔT •  ΔT is always the final temperature minus the initial temperature •  When the temperature increases, ΔT and ΔQ are considered to

be positive and energy flows into the system •  When the temperature decreases, ΔT and ΔQ are considered

to be negative and energy flows out of the system

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Example: A 0.400 kg aluminum teakettle contains 2.00 kg of water at 15.0 °C. How much heat is required to raise the temperature of the water (and kettle) to 100 °C?

( )( )( ) kJ. 712C 85C kJ/kg 186.4kg 2wwww =°°=Δ= TcmQ

The heat needed to raise the temperature of the water to Tf is

( )( )( ) kJ. 6.30C 85C kJ/kg 900.0kg 4.0AlAlAlAl =°°=Δ= TcmQ

The heat needed to raise the temperature of the aluminum to Tf is

Then Qtotal = Qw + QAl = 732 kJ.

Consequences of Different Specific Heats

• Water has a high specific heat compared to land

• On a hot day, the air above the land warms faster

• The warmer air flows upward and cooler air moves toward the beach

CkgJcCkgJc

OH

Si

4186

700

2=

=

What happens at night?

Question

What happens at night?

1. same 2. opposite 3. nothing 4. none of the above

How to determine specific heat?

Calorimeter • A technique for determining the specific heat of a

substance is called calorimetry • A calorimeter is a vessel that is a good insulator that

allows a thermal equilibrium to be achieved between substances without any energy loss to the environment

Calorimetry • Analysis performed using a calorimeter • Conservation of energy applies to the isolated system • The energy that leaves the warmer substance equals the energy that enters the water • Qcold = -Qhot • Negative sign keeps consistency in the sign convention of ΔT

Example: A 0.010-kg piece of unknown metal heated to 100°C and dropped into the bucket containing 0.5 kg of water at 20°C. Determine specific heat of metal if the final temperature of the system is 50°C

Given: Mass: m1=0.010 kg

m2=0.5 kg Specific heat (water): cW =4186 J/kg°C Temperatures:

T1=100 °C T2=20 °C

Tf=50 °C Find: Specific heat =?

( ) ( ) ( )( )( )( ) 0627905.0

205041865.01005001.0

0222

=+−=

−+−=

Δ+Δ==+

JcCCCkgJkgCCckg

TcmTcmQQ

metal

metal

OHOHOHmetalmetalmetalmetalwater

 

Conservation of energy: heat lost by metal is the same as heat acquired by water:

Solve this equation:

0=+ metalwater QQ

CkgJcmetal51025.1 ×=

iron

13

Specific Heat of Ideal Gases

The average kinetic energy of a molecule in an ideal gas is

.23

tr kTK =

And the total kinetic energy of the gas is

.23

23

tr nRTNkTK ==

14

Define the molar specific heat at constant volume; this is the heat capacity per mole.

TnQCΔ

=V

Heat is allowed to flow into a gas, but the gas is not allowed to expand. If the gas is ideal and monatomic, the heat goes into increasing the average kinetic energy of the particles.

Then, the amount of added heat is .23

tr TnRQK Δ==Δ

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In this situation we can calculate the heat capacity with constant volume for ideal gas:

J/K/mol 5.12232

3

V ==Δ

Δ=

Δ= R

Tn

TnR

TnQC

If the gas is diatomic: J/K/mol 8.2025

V == RC

16

Internal energy will be distributed equally among all possible degrees of freedom (equipartition of energy). Each degree of freedom contributes ½kT of energy per molecule and ½R to the molar specific heat at constant volume.

Rotational motions of a 2-atom molecule:

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Example: A container of nitrogen gas (N2) at 23 °C contains 425 L at a pressure of 3.5 atm. If 26.6 kJ of heat are added to the container, what will be the new temperature of the gas?

For a diatomic gas, .V TnCQ Δ=

The number of moles n is given by the ideal gas law .i

ii

RTVPn =

( )( )( )( )( )

K 21/Lm 10L 425/atmN/m 10013.1atm 5.3

K 2965.2

J 106.263325

3

ii

i

V

=

×⎟⎟⎠

⎞⎜⎜⎝

⎛ ×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ

−

RR

VPRT

CQT

The change in temperature is

The final temperature of the gas is Tf = Ti + ΔT = 317 K = 44 °C.

ICE WATER STEAM

Add heat

Add heat

These are three states of matter (plasma is another one)

Phase Transitions

Phase Changes

• A phase change occurs when the physical characteristics of the substance change from one form to another

• Common phases changes are •  Solid to liquid – melting •  Liquid to gas – boiling

• Phases changes involve a change in the internal energy, but no change in temperature

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Phase Transitions

Latent Heat •  During a phase change, the amount of heat is given as

Q = m L

•  L is the latent heat of the substance. Latent heat is the amount of heat per unit mass required to change the phase of a substance. The energy is used to form/break chemical bonds. •  Latent means hidden or concealed

•  Choose a positive sign if you are adding energy to the system and a negative sign if energy is being removed from the system

•  Latent heat of fusion (Lf) is used for melting or freezing •  Latent heat of vaporization (Lv) is used for boiling or

condensing

Problem-solving hints: • Use consistent units • Transfers in energy are given as Q=mcΔT for processes with no phase changes

• Use Q = m Lf or Q = m Lv if there is a phase change •  In Qcold = - Qhot be careful of sign, ΔT = Tf - Ti

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Example: A 75 g cube of ice at -10.0 °C is placed in 0.500 kg of water at 50.0 °C in an insulating container so that no heat is lost to the environment. Will the ice melt completely? What will be the final temperature of this system?

The heat required to completely melt the ice is

( )( )( ) ( )( )kJ 27

kJ/kg 7.333kg 075.0C10C kJ/kg 1.2kg 075.0ficeiceiceiceice

=

+°°=

+Δ= LmTcmQ

The heat required to cool the water to the freezing point is

( )( )( )kJ 105

C50C kJ/kg 186.4kg 5.0wwww

=

°°=

Δ= TcmQ

Since Qice < Qwater the ice will completely melt.

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What will be the final temperature of this system?

To find the final temperature of the system, note that no heat is lost to the environment; the heat lost by the water is gained by the ice.

0 =Qice +Qw

0 =miceciceΔT +miceLf +micecw Tf −Tice,i( )+mwcw Tf −Tw,i( )

0 =miceciceΔT +miceLf + mice +mw( )cwTf −mwcwTi

0 = 27 kJ+ mice +mw( )cwTf −105 kJTf = 32.4 °C

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Example: Compute the heat of fusion of a substance from these data: 31.15 kJ will change 0.500 kg of the solid at 21 °C to liquid at 327 °C, the melting point. The specific heat of the solid is 0.129 kJ/kg K.

kJ/kg 8.22f

f

=Δ−

=

+Δ=

mTmcQL

mLTmcQ

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On a phase diagram, the triple point is the set of P and T where all three phases can coexist in equilibrium.

Sublimation is the process by which a solid transitions into a gas (and gas → solid).

Phase Diagram

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The critical point marks the end of the vapor pressure curve. A path around this point (i.e. the path does not cross the curve) does not result in a phase transition. Past the critical point it is not possible to distinguish between the liquid and gas phases.

Phase Diagram