This Week. 1/22/2018 Physics 214 Summer

This Week

Heat and Temperature

Water and Ice

Our world would be different if water didn’t

expand

Engines

We can’t use all the energy!

Why is a diesel engine more efficient?

Heat

Heat is a form of energy and any object has internal energy in the form of kinetic energy of the atoms or molecules and in potential energy connected with the molecular structure and the electromagnetic forces among the constituents.

If we add energy to an object the internal energy increases.

Energy can be added in many ways. For example we can do work such as

the frictional forces on a moving object.

electromagnetic radiation from the sun.

First Law of Thermodynamics

The increase in the internal energy of a system is equal to the amount of heat added to the system minus the amount of work done by the system.

U = internal energy

Q = heat that is added or removed

W_system = is the work done by the system

Compression the force pushing on the piston does positive work and puts energy into the gas.

The work done by the gas is negative

Expansion The gas pushing on the piston does positive work and energy is removed from the gas The work done by the external F is negative

ENERGY CONSERVATION

∆U = Q – W_system or ∆U = Q + W_external

Compression

Expansion

U

Temperature

Heat flows from a hot body to a cold body until they reach thermal equilibrium.

Temperature is the quantity that measures whether one body is hotter than another and at thermal equilibrium both bodies have the same temperature.

The simplest picture

is a gas of free molecules where the energy is kinetic. A higher temperaturemeans higher velocities and more stored energy.

Increase U and T increases

Measuring temperature

Physical properties of an object change with temperature. For example mercury expands as the temperature

increases and we can use that expansion to measure T.

There are three temperature scales,

Celsius, Fahrenheit and Kelvin

Mixture of ice and water 00_{C 32}0_{F 273}0_K

Boiling point of water 1000_{C 212}0_{F 373}0_K

T_c = 5/9(T_f – 32) T_f = 9/5T_c + 32 T_K = T_c + 273

Note that the temperature at which water boils depends on the value of the atmospheric pressure so that at

Absolute zero

An ideal gas obeys the law PV = constant x T

if we keep V constant the pressure is proportional to T.

If we measure P versus T at constant volume we find a temperature where the pressure is zero. This is absolute zero where the stored internal energy has it’s lowest possible value.

Heat capacity

For a given object at a specific temperature the total internal energy depends on how big the object is. If we add a fixed amount of energy the temperature will rise but the rise will be smaller the larger the object is. So we define a quantity called specific heat

c = is the quantity of heat required to raise unit mass of a substance by one degree.

Q = mc∆T New unit

1 calorie = heat required to raise 1 gram of water 10_{C 1calorie = 4.186 joules}

Note to raise 1kg of water 10_{C requires 1000cal}

Note that internal energy can change without a temperature change. This is due to internal rearrangement of the molecules

Water and Ice

temperature is changed. As we lower the temperature from room temperature it shrinks but it 40_{C it starts to expand and}

that is why ice floats.

Remember density ρ = mass/volume so below 40_{C the volume increases and the}

density decreases.

Above 40_{C the molecules are tightly}

packed. Below 40_{C the molecules form a}

Change of state

We are used to solids, liquids and gases and that most substances can be in any of the three states depending on temperature and pressure. Solids have more orderly structure and the

atoms/molecules are tightly joined. In a liquid the atoms/molecules are loosely joined. In a gas the atoms/molecules are “free”. When we tear apart something that is joined energy is required and this is true as we go from solid to liquid and liquid to gas. Taking water as an example

80 calories/gram are required to melt ice at 0o_{C to water at 0}o_{C and}

540 calories/gram to convert water to gas at 100o_{C. When water}

freezes 80cal/gm has to be removed and when water condenses 540 cals/gm is released.

80 cals/gm is the latent heat of fusion

540 cals/gm is the latent heat of vaporization

Gases

A quantity of gas can be described by three

variables. Pressure, volume and temperature

U = internal energy (kinetic energy of the molecules)

U determines the temperature

Q = heat energy that is put in or taken out

W

= work done by the gas when it expands

or is compressed

W

is + if it expands – if it is compressed

Work done by gas = -Fd = -PAd ∆V = Ad W = -P∆V

Physical Laws for a gas

Energy conservation

∆U = Q + W

(or ∆U = Q - W

)

Ideal gas law PV = NkT (T in degrees Kelvin)

Adiabatic no heat in or out Q = 0

Compression external work is + (∆U, T increase) Expansion external work is – (∆U, T decrease)

Isothermal T does not change ∆U = 0 Q = W = P∆V

put in heat gas expands take out heat gas must be compressed

Isobaric pressure is kept constant

3E-03 Fire Syringe

RAPID COMPRESSION IS ADIABATIC GIVING RAPID RISE OF AIR TEMPERATURE IN THE CHAMBER WHICH EXCEEDS THE IGNITION TEMPERATURE OF THE FLAMMABLE MATERIAL.

Compression and rise in air temperature

This system is analogous to the combustion cycle within a diesel engine or any fuel injected engine.

Can you guess the every-day application of

this

phenomenon ?

Solids and liquids

Temperature depends on the internal energy Put in heat T rises except at a change of state

We define the specific heat c as the amount of heat

required to raise unit mass of a substance by one degree. Q = mc∆T

The usual unit is calories/gram/0_{C 1 cal = 4.186 joules}

The specific heat of water is 1cal/gm/0_C Change of state

ice water steam

Requires 80 cal/gm to melt ice at 0_{C to water at} 0_C

Requires 540cal/gm to make steam at 1000_C

To condense steam or freeze water requires the removal of 540 cal/gm or 80cal/gm.

Q = mc_ice∆T Q = m80 Q = mc_water∆T Q = m540

-T 0 100 Steam

Heat engines

The use of energy and the conversion of energy is

essential not only in our practical everyday life but is a requirement for life to exist. The sun puts energy into the earth and we use such energy sources as gasoline, coal and nuclear power.

The conversion of one form of energy into another form of energy is governed by the physical laws. The generic term used is a heat engine for any system that takes an energy source and uses it to produce work.

A car engine is a practical example.

Of course there is a practical aspect that an energy source can become depleted, like oil but we will find that there is a fundamental law that prevents the

2

nd

_{Law of Thermodynamics}

Heat engine uses energy Q_H and produces work W and releases Q_c

Efficiency = ε = W/Q_H

W = Q_H – Q_c (c = environment)

2nd _law

Carnot Cycle

We can design a “perfect engine”

without friction. The Carnot cycle is is an ideal engine and an analysis

reveals that

ε

For example in a gasoline engine T_H is the temperature of combustion and Tc is outside temperature. If we take T_H as 1500o _{and T}

c as 300o then

Car engines

In an engine with spark plugs a gasoline/air mixture is compressed and the temperature rises and then the spark ignites the mixture before it reaches the combustion temperature. For increased

efficiency one would like to reach as high a temperature as possible so fuel injection solves this problem by compressing air and injecting the fuel after the combustion temperature is reached.

Turbo charged engines force more air into the chamber when the piston reaches it’s lowest point using power from the engine. Super chargers do the same thing but with an extra electrical motor. This causes the engine to run hotter.

Review Chapters 10 and 11

W = P∆V for the piston shown W_external = + W_system =

-∆U = Q – W_system or ∆U = Q + W_external

Temperature scales Celsius, Fahrenheit Kelvin Mixture of ice and water 00_{C 32}0_{F 273}0_K

Boiling point of water 1000_{C 212}0_{F 373}0_K

T_c = 5/9(T_f – 32) T_f = 9/5T_c + 32 T_K = T_c + 273

c = is the quantity of heat required to raise unit mass of a substance by one degree. Q = mc∆T

1 calorie = heat required to raise 1 gram of water 10_{C = 4.186 joules}

Change of state (internal energy changes temperature is constant) 80 cals/gm is the latent heat of fusion

540 cals/gm is the latent heat of vaporization

Work done on gas = Fd = PAd ∆V = Ad W = P∆V

Gases

Energy conservation ∆U = Q – W_gas

Ideal gas law PV = NkT (T in degrees Kelvin) Adiabatic no heat in or out Q = 0

Compression work is + (∆U, T increase) Expansion work is – (∆U, T decrease)

Isothermal T does not change ∆U = 0 Q = W = P∆V

put in heat gas expands take out heat gas must be compressed

Isobaric pressure is kept constant

Put in heat T increases gas expands (hot air balloon)

Work done by gas = -Fd = -PAd ∆V = Ad W = -P∆V

Heat engines

Efficiency = ε = W/Q_H

Change in internal energy in one cycle is zero W = Q_H – Q_c (c = environment)

2nd _law

No engine working in a continuous cycle can take heat at a single temperature and convert that heat completely to work.

Perfect engine (no friction )

ε = (T

–T

)/T

(T in

_K)

T

Geysers (Old Faithful)

Geysers produce a jet of water very often at equal time intervals. Old Faithful in

Yellowstone erupts every 90 minutes or so. What is required is a source of heat, a source of water and a constricted vertical channel that the water flows into.

As the column of water increases in height the pressure increases at the bottom of the water column where the heat is. The increase in pressure raises the boiling point. This

continues until the water at the bottom starts to boil.

This causes an expansion of the column which reduces the pressure and

immediately takes the temperature of all the water below the boiling point and there is an explosive boiling resulting in the eruption.

Questions Chapter 10

Q1

Q2

Celsius degrees?

Water freezes at 0o _{C and 32}o _{F so 0}o _{F is colder}

There is 100o _{C between water freezing and boiling and}

Q4

Q5

Yes. Absolute zero is – 2730 _{which one can think of as the place}

where the molecules of an ideal gas have zero energy

Q6

Q8

temperature of either object change?

They will exchange heat until they both reach the same temperature

Q10

materials, are initially at the same temperature. Equal amounts of heat are added to each object. Will the final temperature of the two objects necessarily be the same?

No. The specific heat, which is the heat energy required to raise the temperature one degree, is different for each material

Q13

Q21

Q23

The temperature will increase ∆U = W

Q25

expand. Will this increased volume of air cause the balloon to fall?

Q27

Archimedes principle states that the buoyant force is equal to the weight of liquid displaced. So if the balloon stays the same size and as the air expands it leaves the balloon it will rise faster

because the weight of the air inside will be less. If no air escapes but the balloon increases in size it will also rise because the

buoyant force is larger

Ch 10 E 2

Temperature is 14° F.

What is the temperature in Celsius?

Ch 10 E6

How much heat does it take to raise the temperature

of 70 g of H

O from 20°C to 80°C? C= 1 cal/gram/

_C.

Ch 10 E 12

Add 600 J to 50 g of H

O initially at 20°C

a) How many calories?

b) What is the final temperature of the H

O

a) 600 /4.186

= 143.3 cal = Q

b) Q=mc

T

T=Q/mc=143.3/(50)(1)=2.87°C

Ch 10 E 16

Add 500 cal of heat to gas. Gas does 500 J of

work on surroundings.

What is the change in internal energy of gas?

U = Q-W

Q = 500/ 4.186

= 2093 J

W=500 J

Ch 10 CP 2

A student’s temperature scale = 0°s = ice point of H₂O; 50° s = boiling point of H₂O. The student then measures a temperature of 15°s.

a) What is this temperature in degrees Celsius? b) What is this temperature in degrees Farenheit? c) What is this temperature in Kelvins?

d) Is the temperature range spanned by 1°s larger or smaller than spanned by 1°C? a) Ice point H₂O = 0°s=0°C Boiling point H₂O = 50°s=100°C Obvious relationship: T_s=1/2 T_c, T_c=2T_s T_c=2T_s=2(15) = 30°C b) T_F = 9/5 T_c+ 32 = 9/5(30) + 32 = 86°F c) T_k = T_c + 273.2 = 30 + 273.2 = 303.2K

d) 1°s spans 2°C, so that the range spanned by 1°S is LARGER than that spanned by 1°C.

T_s T_c

50°s 100°c

Ch 10 CP 4

150 g of metal at 120°C is dropped in a beaker containing 100 g H₂O at 20°C. (Ignore the beaker). The final temperature of metal and water

is 35°C.

a) How much heat is transferred to the H2O? b) What is the specific heat capacity of metal?

c) Use the same experimental setup – how much metal at 120°C to have final temperature of metal and water = 70°C?

Ch 11 E 6

A Carnot engine takes in heat at 650 K and releases

heat to a reservoir at 350K.

What is the efficiency?

Ch 11 E 8

Carnot engine operates b/w 600 K and 400 K and

does 200 J of work in each cycle.

a) What is the efficiency?

b) How much heat does it take in from the high-temp

reservoir during each cycle?

a) ε

= (T

- T

/T

= (600-400)/600

= 0.33

Ch 11 CP 2

Carnot engine operates b/w 500° C and 150° C and does

30 J of work in each cycle.

a) What is the efficiency?

b) How much heat is taking in from the high-temp reservoir each cycle? c) How much heat is released to low-temp reservoir each cycle?