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Thermodynamics. Chapter 15. Calculating Work. Work = area under Pressure vs. Volume graph W = Fd F = PA W=PAd W = P D V Calculus link W = - p dV. V 2. V 1. Isochoric (isovolumetric). No change in volume W = 0. Isobaric Process . No change in pressure

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thermodynamics

Thermodynamics

Chapter 15

calculating work
Calculating Work

Work = area under Pressure vs. Volume graph

W = Fd

F = PA

W=PAd

W = PDV

Calculus link

W = - p dV

V2

V1

isochoric isovolumetric
Isochoric (isovolumetric)
  • No change in volume
  • W = 0
isobaric process
Isobaric Process
  • No change in pressure
  • Volume expands and does work
  • WORK = - AREA
work from graph example 1
Work from Graph: Example 1

A gas expands at a constant pressure of 1 X 105 N/m2 (~100 kPa, ~ 1 atm) from a volume of 10 mL to a volume of 35 mL. Calculate the work done by the gas.

slide7
W = PDV

W = (1 X 105 N/m2)(0.035 L – 0.010 L)

W = 2500 J

work from graph example 2
Work from Graph: Example 2

A gas is cooled to produce a decrease in pressure. The volume is held constant at 10.0 m3, and the pressure decreases from 1 X 105 N/m2 to 0.5 X 105 N/m2. Calculate the work done on the gas.

slide10
W = PDV

W = P X 0

W = 0 J (No work is done)

work from graph example 3
Work from Graph: Example 3

A gas expands and increases in pressure as shown in the next graph. Calculate the work done on the gas. (Remember to include the entire area under the graph).

example 4
Example 4

Calculate the work done from 500 to 1000 cm3 on the following graph. Remember to go all the way down to the x-axis.

isothermal systems
Isothermal Systems
  • Area under the curve

W = - p dV = - nRT dV = - nRT dV

V V

W = -nRT ln Vf = -piVi ln Vf = pfVf ln Vf

Vi Vi Vi

isothermal example
Isothermal Example

A cylinder contains 7.0 grams of nitrogen (N2).

  • Calculate the moles of N2
  • Calculate the work that must be done to compress the gas at a constant temperature of 80oC until the volume is halved.
slide18

Top Line = at 500K

Bottom Line = at 300K

work and cyclic processes
Work and Cyclic Processes
  • Cyclic process – gas returns to its original state
  • Important for studying
    • Steam engine
    • Car engine

DU = 0

Q = W

  • Work = Area enclosed
work and cyclic processes1
Work and Cyclic Processes

Calculate the work shown in the previous graph.

heat q
Heat (Q)
  • Heat – Energy transferred from one body to another because of a difference in temperature
  • Cooking a turkey

hot oven  cooler turkey

  • Extensive Property – depends on amount of material (iceberg vs. water)
  • Unit
    • Joules.
    • Calorie
calories
calories
  • calorie – amount of heat energy needed to raise the temperature of 1 gram of water by 1 degree Celsius (or Kelvin)
  • Not a nutritional Calorie

1 nutritional Calorie = 1000 calories (or 1 kilocalorie)

converting between heat units
Converting between heat units

1 cal = 4.18 Joules

252 cal = 1 BTU

1054 Joules = 1 BTU

james prescott joule
James Prescott Joule
  • Weight moves paddles
  • Friction from paddles warms water
  • Workfalling = Heatpaddles
  • Mechanical Equivalent of Heat
heat example 1
Heat: Example 1

How high would you have to climb to work off a 500 Calorie ( 500,000 cal) ice cream? Assume you mass 60 kg.

(500,000 cal)(4.186J/cal) = 2.09 X 106 J

Heat = Work

Heat = mgh

h = Heat/mg

h = 2.09 X 106 J/(60 kg)(9.8 m/s2) = 3600 m

h ~ 11,000 ft

heat example 2
Heat: Example 2

A 3.0 gram bullet travels at 400 m/s through a tree. After passing through the tree, the bullet is now only going 200 m/s. How much heat was transferred to the tree?

slide29
KE = ½ mv2 KE = ½ mv2

KE = ½ (0.003 kg)(400 m/s)2 KE = ½ (0.003 kg)(200 m/s)2

KE = 240 J KE = 60 J

Heat loss = 180 J to the tree

the laws of thermodynamics
The Laws of Thermodynamics

1st Law

  • Energy is conserved

DE = W + Q

2nd Law

  • Natural processes tend to move toward a state of greater disorder
  • Heat goes hot to cold (Clausius)
  • No device converts all heat to work (Kelvin-Planck)

DS >0

the laws of thermodynamics1
The Laws of Thermodynamics

3rd Law

  • The entropy of a pure crystal at absolute zero is zero

DG = DH - TDS

important definitions
Important Definitions

Thermodynamics – Study of the transfer of energy as heat and work

System – Objects we are studying

first law sign conventions
First Law Sign Conventions

Heat is added to the system +

Heat is lost -

Work on the system +

Work done by the system -

first law example 1
First Law Example 1

2500 J of heat is added to a system. This heat does 1800 J of work on the system. Calculate the change in internal energy.

DE = W + Q

DE = 1800 J +2500 J

DE = 4300 J

first law example 2
First Law Example 2

2500 J of heat is added to a system. This increase in temperature allows the system to do 1800 J of work. Calculate the change in internal energy.

DE = W + Q

DE = -1800 J +2500 J

DE = 700 J

adiabatic systems
Adiabatic Systems

– No heat exchange (Q=0)

  • Fast processes
  • Heat has no time to enter/leave system
  • Car piston

Q = 0 For an ideal gas

DE = W + Q DE = 3/2 nRT

DE = W DE = 3nRDT

2

work example 1
Work: Example 1

In an engine, 0.25 moles of gas in the cylinder expand adiabatically against the piston. The temperature drops from 1150 K to 400K. How much work does the gas do? (the pressure is not constant).

slide38
Q = 0

DE = Q-W

DE = W

DE = 3nRDT

2

DE = 3(0.25 mole)(8.315 J/K-mol)(400 K - 1150 K)

2

DU = 2300 J

W = 2300 J

temperature and internal energy
Temperature and Internal Energy

Temperature – measure of the average kinetic energy of individual molecules

Internal (Thermal) Energy – Total energy of all the molecules in an object

U = 3 nRT

2

temperature
Temperature

Suppose we heat mugs of water from 25oC to 90oC.

One Mug Two Mugs

25oC to 90oC 25oC to 90oC

Same Temp. Change Same Temp. Change

Requires less heat Requires more heat

specific heat
Specific Heat

Specific Heat – Amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius or Kelvin

  • Unit – J/kgoC
  • Symbol = C
  • The higher the specific heat, the more energy needed to raise the temperature
  • Wooden spoon versus a metal spoon
calculating heat
Calculating Heat

Examples:

  • Al has a specific heat of 0.22 kcal/kg oC
  • Gold has a specific heat of 0.03 kcal/kg oC
  • Which gets hotter sitting in the sun?
  • Water’s specific heat: 1.00 kcal/kg oC or 4186 J/kgoC
calculating heat1
Calculating Heat

Suppose you immerse a hot pan into a dishpan of water to cool it. Which will experience a greater change in temperature (gain or loss)?

Heat lost = -Heat gained

Q1 + Q2 + Q3 +…. = 0

calculating heat2
Calculating Heat

Q = mCpDT

Q = heat (J)

m = mass (kg)

Cp = specific heat (J/kgoC)

DT = Tfinal – Tinitial

calculating heat example 1
Calculating Heat: Example 1

How much heat must be supplied to a 500.0 gram iron pan (C = 450 J/kgoC) to raise its temperature from 20.0oC to 100oC?

Q = mCDT

Q = (0.500 kg)(450/kg oC)(100oC -20oC)

Q = (0.500 kg)(450/kg oC)( 80oC)

Q = 18,000 J or 18 kJ

calculating heat example 2
Calculating Heat: Example 2

Suppose the pan is filled with 400 g of water. What would be the total heat?

Q = mCDT

Q = (0.400 kg)(4186/kg oC)(100oC -20oC)

Qwater =134,000 J

Qtotal = 152,000 J or 152 kJ

calculating heat example 3
Calculating Heat: Example 3

200-g of tea at 95oC is poured into a 150-g glass (C = 840 J/kgoC) at 25oC. What will be the final temperature of the cup/glass?

Heat lost = -Heat gained

Qtea = - Qcup

mteaCteaDT = -mcupCcupDT

slide48
mteaCteaDT = -mcupCcupDT

(0.200 kg)(4186J/kgoC)(T-95oC) =

-(0.150 kg)(840J/kgoC)(T-25oC)

(837)(T-95oC) = -(126)(T-25oC)

837T – 79,500 = 3150 -126T

963 T = 82,700

T = 86oC

calculating heat example 4
Calculating Heat: Example 4

A 100-g piece of aluminum (C = 900 J/kgoC) is heated to 100 oC and immersed in 250-g of water at 20 oC. What is the final temperature of the system?

Heat lost = -Heat gained

Qtea = - Qcup

mAlCAlDT = -mwaterCwaterDT

slide50
mAlCAlDT = -mwaterCwaterDT

(0.100 kg)(900J/kgoC)(T-100oC) =

-(0.250 kg)(4186J/kgoC)(T-20oC)

(90)(T-100) = -(1047)(T-25)

90T – 9,000 = 20,930 - 1047T

1137T = 29,930

T = 26oC

phase changes
Phase Changes

Definition - A change of state in which energy is absorbed or released without a temperature change.

Example: Freezing water

  • Water is cooled from 25oC to 0oC (temperature change, not a phase change)
  • Water freezes at 0oC (no temperature change occurs during freezing, phase change)
phase change
Phase Change
  • Suppose we are heating ice:
  • All of the heat goes into melting the ice rather than increasing the temperature
  • Temperature will only rise once it is all melted

Solid molecules absorb the heat to get moving rather than increasing temperature

Melted molecules now move freely

phase changes think about it
Phase Changes: Think about it.
  • Can you get water above 100oC in a pot if you are cooking?
  • Can you get steam above 100oC in a pressure cooker or furnace?
  • Can you get ice below 0oC in a freezer?
heating curves
Heating Curves

Boiling

Melting

latent heat
Latent Heat

No Phase Change

Q = mCpDT

Phase Change

Q = mLfor Q = mLv

Lf = Latent heat of fusion (freezing/melting)

Lv = Latent heat of vaporization (boiling/condensing)

heating curves1
Heating Curves

Temperature (oC)

Use q = mLv

Boiling

Use q = mLf

Melting

Use q = mCpDT

Heat (Joules)

important values
Important Values

Here are some important values that we will need:

Substance C

Steam 2010 J/kg oC

Water 4186 J/kg oC

Ice 2100 J/kg oC

Latent Heat of fusion (water) Lf = 3.33 X 105 J/kg

Latent Heat of vaporization(water) Lv = 22.6 X 105 J/kg

heating curves example 1
Heating Curves: Example 1

How much energy is required to cool 100.0 grams of water at 20.0oC to make ice at –10.0oC?

There are three steps involved here:

  • Cooling the water
  • Freezing the water
  • Cooling the ice down to –10.0oC

Let’s look at the graph:

slide59

Temperature (oC)

Freezing (q=mLf)

Water cools (q=mCpDT)

Ice cools (q=mCpDT)

Heat (Joules)

slide60
Let’s look at it step by step
  • Cooling the water

Q = mCpDT = (0.100 kg)(4186J/kgoC)(0oC-20oC)

Q = 837 J

  • Freezing the water

Q = mLf = (0.100 g)(3.33 X 105 J/kg)= 3.33 X 104 J

slide61
Cooling the ice down to –10.0oC

q = mCpDT = (0.100 kg)(2100J/kgoC)(-10oC-0oC)

q = 2100 J

Lastly, we just add the three heat values together:

837J + 33,000J + 2100J = 36,200 J

heating curves example 2
Heating Curves: Example 2

How much energy does a refrigerator have to remove from 1.5 kg of water at 20 oC to make ice at -12 oC?

ANS: 660 kJ

calorimeter
Calorimeter
  • Foam cup
  • Insulated metal container (thermos)
  • Must consider heat absorbed by the calorimeter in the equations
calorimeter example 1
Calorimeter: Example 1

A 0.150-kg sample of a new alloy is heated to 540 oC. It is placed into 400-g of water at 10.0 oC which is contained in a 200-g Aluminum calorimeter. The final temperature of the mixture is 30.5 oC. Calculate the specific heat of the new alloy.

slide65
Heat lost = - Heat gained

malloyCalloyDT = -[mwaterCwaterDT + mcalCcalDT]

mwaterCwaterDT = (0.400kg)(4186 J/kg oC)(30.5 oC – 10.0oC)

mwaterCwaterDT = 34,300 J

mcalCcalDT = (0.200kg)(900 J/kg oC)(30.5 oC – 10.0oC)

mcalCcalDT = 3700 J

slide66
malloyCalloyDT = -[mwaterCwaterDT + mcalCcalDT]

malloyCalloyDT = -[34,300 J + 3700 J]

malloyCalloyDT = -38,000 J

(0.150-kg)(Calloy)(10 oC – 540 oC) = -38,000 J

(-79.5 kg oC)(Calloy) = -38,000 J

Calloy = 478 J/ kg oC

example 2
Example 2

100 g of ice at -20oC is added to 500 g of soad at 20oC.

  • Calculate the heat required to raise the temperature of the ice to the melting point.
  • Calculate the heat required to melt the ice.
  • Is there enough heat energy available if the soda cools to 0oC?
  • Calculate the final temperature of the soda
heat flow
Heat Flow
  • Conduction
  • Convection
  • Radiation
conduction
Conduction

Conduction – The transfer of energy from molecule to molecule through collisions.

  • Warmer (faster) molecules to slower(colder) molecules
  • Thermometers work through conduction.
  • Depends on the area of contact.
conduction thermometer
Conduction: thermometer

Slower (colder) mercury atoms get “bumped” and accelerated by the collisions with the air

The warmer, faster moving air molecules collide with the glass and give some of their kinetic energy of motion to the glass and mercury.

conduction1
Conduction
  • Air is a poor conductor - molecules are so far apart, far fewer collisions than in a solid.
  • Insulators usually have a number of air pockets (like in home insulation) to slow the heat flow.
  • Cooking on top of a stove is usually conduction since the pan/pot is in contact with a hotter burner element or a flame.
conduction2
Conduction

DQ = kA(T1 – T2)

Dt L

DQ= Rate of heat flow (heat/time)

Dt

k = Thermal Conductivity

A = Area

T1 – T2 = Change in Temperature

L = Length

conduction example 1
Conduction: Example 1

Calculate the rate of heat flow through a window 3 m2 and 3.2 mm thick. The outer and inner temperatures are 14.0 oC and 15.0 oC

conduction r values
Conduction: R-values
  • R values – commercial measure of the quality of insulation
  • Higher R = more insulation
  • R = length

k

R

Glass 1

Brick 1

Insulation 12-20

convection
Convection

Convection – the transfer of hear by the movement of a large volume of air or liquid

  • Hot-air home heating
  • Radiator wars air that it comes in contact with through conduction. Convection takes over as it moves through the room
  • Convection ovens
convection1
Convection

Then convection causes the warm air to expand into the room (the hot air rises)

A hot radiator warms air through conduction (contact between the atoms/ molecules

The cold air sinks and comes in contact with the radiator again.

radiation
Radiation

Radiation – The transfer of heat through electromagnetic waves

  • Light (microwaves, infrared, visible, etc..)
  • The waves give kinetic energy to molecules and cause them to move faster (increase in temperature).
  • Sun
  • Red heat lamps at fast-food places.
radiation1
Radiation

Light and other radiation from the sun strike the earth impart some of their energy to the molecules on the earth

Earth

radiation2
Radiation

Stefan-Boltzmann Equation

DQ = esA(T14 – T24)

Dt

s = 5.67 X 10-8 W/m2 K4 (Stefan-Boltzmann constant)

A = area

T = Temperature (kelvin)

e = Emissivity

radiation emissivity
Radiation: Emissivity

e = Emissivity

  • Value between 0 and 1
  • 1
    • Black substances
    • Absorb and emit radiation well
  • 0
    • Shiny substances
    • Do not absorb or emit well
radiation example 1
Radiation: Example 1

An athlete is sitting in a locker room at is 15 oC. The athlete has a skin temperature of 34 oC and an evalue of 0.7. If his surface area is 1.5 m2, calculate the rate of heat loss.

DQ = esA(T14 – T24)

Dt

DQ = (0.7)(5.67 X 10-8 W/m2 K4)(1.5m2)(3074 – 2884)

Dt

DQ/Dt = 120 W

radiation example 2
Radiation: Example 2

A ceramic teapot (e=0.70) and a shiny teapot (e = 0.10) filled with tea at 95 oC. Calculate the rate of heat loss of each if the room is 20oC. Assume each teapot is about 0.05 m2.

ANS: 22 W, 3.1 W

radiation and sun angle
Radiation and Sun Angle
  • Get less radiation as the sun becomes less vertical
  • Would you absorb more sun at midday or 6 PM?
  • Would you absorb more sun in Mexico or Ohio?
sun example 1
Sun: Example 1

What is the energy absorption of a person (0.80 m2 , e = 0.70) getting a tan if the sun makes an angle of 30o with the vertical.

DQ = (1000 W/m2)eA cos q

Dt

DQ = (1000 W/m2)(0.7)(0.80m2) cos 30o

Dt

DQ/Dt = 490 W (wearing light clothes will reduce e)

thermal pollution
Thermal Pollution
  • Thermally polluting power plants
    • Coal plants
    • Oil plants
    • Nuclear plants
  • Non-thermally polluting power plants
    • Hydroelectric
    • Tidal energy
    • Wind
    • Solar
work example 4
Work: Example 4

Determine the change in internal energy of 1.0 liter of ware at 100oC when it is fully boiled. It results in a 1671 L of steam at 100oC. Assume the process is done at atmospheric pressure.

DU = Q – W

Q = mL =(1.0 kg)(22.6 X 105 J/kg) = 22.6 X 105 J

slide95
W = PDV (1 L = 1 X 10-3 m3)

W = (1 X 10 5 N/m2)(1671 X 10-3 m3 - 1X 10-3 m3)

W = 1.7 X 105 J

DU = Q – W

DU = 22.6 X 105 J – 1.7 X 105 J

DU = 21 X 105 J (or 2.1 X 106 J)

work example 2
Work: Example 2

Determine the change in internal energy of 1.0 liter of ware at 100oC when it is fully boiled. It results in a 1671 L of steam at 100oC. Assume the process is done at atmospheric pressure.

DU = Q – W

Q = mL =(1.0 kg)(22.6 X 105 J/kg) = 22.6 X 105 J

slide97
W = PDV (1 L = 1 X 10-3 m3)

W = (1 X 10 5 N/m2)(1671 X 10-3 m3 - 1X 10-3 m3)

W = 1.7 X 105 J

DU = Q – W

DU = 22.6 X 105 J – 1.7 X 105 J

DU = 21 X 105 J (or 2.1 X 106 J)