1

Natural Sciences I

lecture 8: More on Heat & Temperature; Thermodynamics

In the last lecture, we learned that the temperature of a substance is a measure of the kinetic energy of its constituent molecules. It is clear that HEAT and temperature are related in some way, too – but in this case the relationship is more complex. Heat involves more than just the average kinetic energy of the molecules. Let's begin sorting this out by making a clear distinction between two types of energy that a substance or object can have...

EXTERNAL ENERGY is the total kinetic and potential energy that an object has because of its speed, position, etc.

KE = ½ m v

INTERNAL ENERGY is the total kinetic and potential energy of the molecules that make up the object.

2

Kinetic energy: pulsing, twisting, rotating of molecules or atoms

PEgrav = mgh The external energy does not depend upon the temperature of the object. An interesting example is a meteor streaking toward Earth: the object itself has both kinetic and potential energy...

(-)

(+)

F

F

Potential energy is contained in bonds: attraction of (+) and (-) ions, etc.

Internal energy depends upon:

...but a closer look would reveal that something is missing in the energy inventory if we consider 2 just ½mv and mgh.

temperature density volume pressure (esp. for gases)

2 Here's another way of looking at internal vs. external energy: EXTERNAL ENERGY is kinetic and potential energy you can see. INTERNAL ENERGY is the total kinetic and potential energy of the molecules (which you can't see under normal circumstances). As with all forms of energy, interconversions are possible.

Example: Energy changes associated with pushing a weight up a ramp at a constant velocity...

What is happening, energy-wise? net force

WORK is being done in overcoming friction and gravity frictional heating

The weight has KINETIC ENERGY because it is moving POTENTIAL ENERGY is being created by change of position in Earth's gravity field External mechanical energy is being converted into INTERNAL ENERGY (in this case heat)

Let's return to the difference between HEAT and TEMPERATURE... TEMPERATURE is a measure of the internal kinetic energy only HEAT is a measure of the total internal energy, and thus includes the energy associated with the bonds between atoms Here's "proof" of the difference: water 1 liter o 90 C

250 ml o 90 C

Which beaker of water require o more ice to cool to 40 C? Which would take longer to re-heat o over a burner to 90 C?

Even though the kinetic energy of the molecules is the same, the internal energy of the 1-liter beaker is clearly greater.

3 The preceding example may not be that convincing, because all it shows is that the more molecules you have, the more internal energy (heat) you have. But, what if the molecules are different? water

1 liter o 20 C

alcohol

1 liter o 20 C

The water and alcohol are at the same temperature (equivalent KE of molecules), but... Which would require more ice to cool o 10 C? Which would re-heat faster over a burner?

HEAT AS ENERGY TRANSFER (measurement of heat) Heat is usually measured during a process that involves gain or loss (transfer) of energy from one body to another, or from an energy source to a body. We can thus define heat as... HEAT is a measure of the internal energy that has been absorbed by a body or transferred from one body to another.

DHEAT

=

DENERGY

In general, the process of increasing the internal energy of a body is called "heating" and the process of decreasing the internal energy is called "cooling". The words "heating" and "cooling" thus describe the direction of energy transfer: into or out of a body. Heating and cooling can take place by direct transfer of energy from one body to another (always from hot to cold), or by some sort of energy conversion process, e.g. radiant energy mechanical energy electrical energy

heat heat heat

4

Units of Heat You are already familiar with the joule (J), which is a unit of work, energy or heat 1 calorie (cal) = 4.184 J 1 Calorie (nutritional) = 1 kilocalorie (kcal) 1 British Thermal Unit (Btu) = 252 cal Mechanical equivalents 1 Btu = 778 ft-lb Interesting(?) facts: o

1 cal is the energy needed to raise the temperature of 1 g of water 1 C o

1 Btu is the energy need to raise the temperature of 1 lb of water 1 F Example: Let's try using these units in a work/energy conversion that has some direct relevance to how we eat... If you weigh 120 pounds, how many flights of stairs (12 ft high) would you have to climb in order to "work off" a 240 Calorie candy bar? Since you're given the weight/distance information in pounds and feet, let's start by calculating the work done climbing 1 flight of stairs in ft-lb. (120 lb) x (12 ft) = 1440 ft-lb per flight of stairs Using the equivalents above, we can convert this to Calories... 1440 ft-lb

1 Btu 778 ft-lb

252 cal = 466.4 cal = 0.4664 Calories 1 Btu

This sobering result indicates that you burn only about half a Calorie climbing one flight of stairs. The work off the candy bar, you would need to climb 1 flight = 515 flights of stairs! 240 Calories 0.4664 Calories

5 Another Example: Conversion of mechanical energy into heat... A 1,000 kg automobile is moving at 90 km/h (25 m/s) and suddenly brakes to a stop. How many kilocalories are generated?

hot brakes This is an example of conversion of kinetic (mechanical) energy into heat, so the first thing to do is calculate the kinetic energy of the moving automobile. 2 KE = ½ m v = ½ (1,000 kg) (25 m/s) 2

2

2

= (500) (625) kg m /s = 312,500 J

This can be expressed in kilocalories using the conversion factor on p. 4: 312,500 J

1 kcal 4,184 J

= 74.7 kcal

Let's return to some examples of processes involving heat transfer

10 kg of water, o initially at 20 C

100 Calorie hot dog (= 100,000 calories)

If all of the internal (chemical) potential energy contained in the hot dog went into heating the water, the temperature of the o water would rise by 10 o to 30 C.

6 aluminum block: 454 g o initially at 50 C

If the aluminum block is dropped into the water and allowed to cool o from 50 to 40 C, it will heat the water from 10 o to 11 C.

water: 1 kg (1000 g) o initially at 10 C

What we have really done in the last two experiments is something called CALORIMETRY. We'll return to this after we get some more background on...

SPECIFIC HEAT We just looked at a couple of examples of heat transfer. The temperature changes that occur as one body transfers heat to another actually depend upon two aspects of each body: its MASS and its SPECIFIC HEAT; the latter is a property unique to a given material. Specific heat is the amount of energy required to raise the temperature of 1 g of a substance by 1 degree C (it may be clear from the preceding discussion that the specific heat of water is one calorie – but this is a little circular because this is how a calorie is defined!). Specific heat usually has the symbol c. Examples...

o

o

100 C

20 C 176 cal absorbed 10 g aluminum

o

c = 0.22 cal/g C

176 cal released

10 g copper

74.4 cal absorbed

10 g gold

24 cal absorbed

o

c = 0.093 cal/g C 74.4 cal released

o

20 C

24 cal released

o

c = 0.03 cal/g C o

100 C

7 Specific heat is related to the internal (atomic) structure and bonding. As heat flows into a substance, some of it goes into increasing the internal kinetic energy of the atoms or molecules and some goes into increasing the internal potential energy. Different materials require different amounts of heat (energy) input to raise their temperatures a given amount. The quantity of heat is usually symbolized as Q...

Q = mcDT This equation can be used to perform calculations involving both heating and cooling. Negative DT, would thus yield negative Q – meaning that heat was flowing out of the material experiencing the negative DT. During heat exchange between two materials heat lost from hot material = heat gained by cooler material

Qlost = Qgained mcDTlost = mcDTgained Example calculation: How many kilocalories of heat must be supplied to a 500-g o o aluminum pan to raise its temperature from 20 C to 100 C? m = 500 g

Q = mcDT o

c = 0.22 cal/g C o

o

o

= (500 g) (0.22 cal/g C) (80 C)

DT = +80 C

= (500) (0.22) (80) g

Q=?

= 8,800 cal

cal o gC

o

C

= 8.8 kcal Interestingly, if the pan were iron rather than aluminum (and of equivalent mass – 500 g) only half as much heat would be required, because the specific heat of iron is half that of aluminum.

8 You now have some basic information on heat exchange and specific heat. Up to this point, however, we have discussed the process of heat transfer in a very general way, indicating only that it can occur by direct contact or by radiative transfer. The details of heat flow need further development...

HEAT FLOW Heat transfer due to a temperature difference between two bodies can occur in three different ways:

Conduction Radiation Convection Let's run through these one at a time: CONDUCTION Conduction involves the direct transfer of energy from molecule to molecule of a material or between two materials Example: Heating the end of an iron bar with a flame. The heat is transferred along the bar by conduction...

short time intermediate time

short temperature

long time

intermediate long time

distance

9 Conduction can also occur between one material and another... initial condition (time = 0)

intermediate time

long time

temperature

HOT COLD

distance

Questions: What has been assumed in drawing these pictures and graphs (Hint: Would the two blocks exchange heat only with each other?) What is the eventual end-point of this exchange of heat? What assumptions would you have to make in order to answer this question? Molecular model of conduction: Progression of atomic vibrations through the substance... vibrational amplitude

HOT

COLD

The rate of conduction depends upon the temperature difference and the nature of the material. just in case you're interested...

JE = energy flux =

joules DT = K C Dx 2 cm s

K = thermal conductivity C = heat capacity (similar to specific heat)

10 Materials vary enormously in their ability to conduct heat... (see Table 5.3 of text) material silver copper aluminum iron lead concrete glass tile brick water wood cotton styrofoam glass wool air vacuum

conduction rate (cal/s)* 0.97 0.92 0.50 0.11 0.08 4.0E-3 2.5E-3 1.6E-3 1.5E-3 1.3E-3 3.0E-4 1.8E-4 1.0E-4 9.0E-5 6.0E-5 0

* Based on oa temperature

gradient of 1 C/cm and a 2 cross-sectional area of 1cm metals

"ceramics" (= oxides)

materials with "voids"

no molecules!

Heat transfer by RADIATION We've seen this before in our discussion of radiant energy... HOT

COLD

incident rays induce vibrations of molecules in near-surface region

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Heat transfer by CONVECTION Convection involves physical transport of molecules possessing a large amount of kinetic energy – it is a mechanical process that involves bulk movement of hot material from one place to another...

household convector

hot

coffee mug

cooler

cooler

Examples:

HEAT and PHASE CHANGES Phase changes involve reorganization of the molecular (atomic) structure of a substance, without change in composition. A phase change always absorbs or releases energy (heat) that is not associated with a change in temperature. EXAMPLES: solid solid liquid gas gas

liquid gas gas liquid solid

(melting) (sublimation) (evaporation) (condensation) (condensation)

Each of these types of phase changes is associated with an energy change that we refer to as (for example) a HEAT OF MELTING or HEAT OF VAPORIZATION, etc. [Again, some of you may recognize these "heats" as enthalpies, enthalpy changes, or "DH's". We'll stick with the term "heat" for now...] These energy changes should be intuitively reasonable in the context of what we've already said about the energies of materials. It should make sense to you now that the energy of liquid water must be higher than that of ice at the same temperature, because of the greater kinetic energy possessed by the H2O molecules in water... (next page)

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Phase Changes (cont'd) o

o

The difference in energy between ice at 0 C and water at 0 C is the heat of melting or heat of fusion of H2O. This is commonly referred to as a latent heat, because it is not detectable by a temperature change [We sometimes say it is not "sensible" heat, meaning that our senses cannot detect it. The word "latent" means "hidden"]. This same heat may also be referred to as the heat of crystallization of H2O – it really just depends on your perspective: Are you warming the system from below the melting point, or are you cooling it from above the melting point? Your text uses the symbols Lf and Lv for the latent heats of fusion and vaporization, respectively. The values for H2O are: Lf (H2O) = 80.0 cal/g Lv (H2O) = 540.0 cal/g The energy change associated with melting (or crystallizing) a given mass o of H2O at 0 C is simply Q = m Lf Similarly, the energy change associated with evaporating (or condensing) a given mass of H2O is Q = m Lv Your new knowledge about latent heats can now be used to understand the full range of heat and temperature effects as ice (or any other substance) is gradually heated from a temperature below the melting point all the way to vapor at a temperature above the boiling point. The two things you need to remember are the relationships immediately above and that incorporating specific heat, which was discussed last class: Q = m c DT energy input over a change in temperature

change in temperature (Tf - Ti )

mass of material specific heat of material (your text uses the term "specific heat capacity")

13 heat-temperature graph

boiling

100

burner with constant heat output

40 20 0

g rm in

60

-20

heat input ice + ice water

The specific heats (c's) and latent heats (L's) are given in Table 5.4 of text

phase change

wa

ICE

phase change warming

Temp. sensor

80

melting

temperature (C)

1 atmosphere frictionpressure less

warming

120

physical set-up

water + vapor vapor

water

phases present

Q2 = m L f

Q3 = mcwDT

Q1 = mciDT

Q4 = m L v Q5 = mcvDT

heat input equations

Example calculation: Let's find out just how much energy is required to o

o

heat 100 g of ice at -20 C to water vapor at 120 C. o

o

step 1: heat ice (-20 C - 0 C) o

step 2: melt ice at 0 C o

o

o

o

Q1 = (100g)(0.50 cal/g C)(20 C)

= 1.0 kcal

Q2 = (100g)(80 cal/g)

= 8.0 kcal

o

o

step 3: heat water (0 C - 100 C) Q3 = (100g)(1.0 cal/g C)(100 C) o

step 4: boil water at 100 C o

step 5: heat vapor to 120 C

Q4 = (100g)(540 cal/g) o

= 10.0 kcal = 54.0 kcal

o

Q5 = (100g)(0.48 cal/g C)(20 C)

= 0.96 kcal

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EVAPORATION, CONDENSATION and SATURATION air at atmospheric o pressure and 20 C

?

Question: Are there H2O molecules in the air inside the box?

?

Answer: Yes, because some of the molecules in the water have enough kinetic energy to "escape" at the surface.

water

Conclusion: Evaporation actually takes place all o the time at temperatures well below 100 C. o

The energy required to evaporate at a temperature below 100 C is that o required to heat up to 100 C and then evaporate (so low-temperature evaporation "consumes" more energy than evaporation at the boiling point).

Question: What is the numerical value of the heat of sublimation? EVAPORATION absorbs heat and thus has a cooling effect at the location where it occurs (How much energy is absorbed when 1 g of sweat evaporates from your skin??). Effect of Pressure on Evaporation

high pressure

low pressure

pressure cookers vacuum drying cooking in Denver

CONDENSATION is the reverse of evaporation. In the boxes pictured above, some H2O molecules in the air will collide with the water surface, lose some of their kinetic energy, and become incorporated into the liquid. If the system remains at constant conditions for a long time, an equilibrium is reached between molecules entering the air and molecules returning to the water. CONDENSATION releases heat and can have the effect of maintaining temperature prevents frost damage to crops provides the energy that drives hurricanes

15 On the preceding page, an equilibrium condition was described in which equal numbers of H2O molecules evaporate and condense in any given time. This is referred to as a saturated condition, or simply SATURATION. The number (concentration) of H2O molecules in H2Osaturated air is very sensitive to temperature: The absolute humidity is simply the concentration of H2O molecules in the air -without reference to how much moisture the air is capable of holding.

40

"forbidden" region

ra

te

d

30

a

tu

20 10 0 -20

permissible region

s

"capacity"

H 2 O in air at saturation (g/m3 )

50

0

20

40 o

temperature ( C)

relative humidity =

60

Often of greater interest, (especially if you're a person who is sensitive to humidity) is the RELATIVE HUMIDITY, which is the amount of moisture in the air relative to what the air can hold at its present temperature

water vapor in the air capacity at present temperature 3

g/m (present) R.H. = 3 g/m (maximum)

100 (%)

100 (%)