Heat

Heat

1

Heat is… energy that is transferred from one body to another because of a difference in temperature.

2

units of heat calorie:

the amount of heat necessary to raise the temperature of 1 gram of water by 1 Celsius degree.

Calorie:

(with a capital C) also known as kilocalorie (kcal), is the heat needed to raise 1 kg of water by 10

Celsius. Btu: temperature of F.

British thermal unit is the heat required to raise the 1 lb of water by 10 equal to 0.252 kcal = 1055 J

3

A given amount of work is always equivalent to a particular amount of heat input: 

4.186 J = 1 cal

4.186 x 10

3

J = 1 kcal

4

Kinetic energy transformed to heat (energy conservation): KEi = KEf + Q where: KEi = KEf = Q =

Initial kinetic energy final kinetic energy heat produced

5

Examples: •How many calories are required to raise the temperature of 8 grams of water by 50 C? Solution: No. of cal = 8 grams x 1 cal/gram-10C (50 C) = 40 cal Answer 2. A couple ate too much ice cream and cake. They each overrated by 500 Calories. To compensate they wanted to (a) do an equivalent amount of work by climbing stairs. (b) How much total height must each climb if each had a mass of 60 kg? Solution:

500 Calories (a)

answer

(b)

500 kcal

Joules?

W = 500 kcal (4.186 x103 J/kcal) = 2.1 x 106 J from W = mgh, h = W/mg h = (2.1 x 106) J/ 60 kg x 9.8 m/s2 = 3600 m

(!!)

6

…examples

3. When a 3.0-g bullet, travelling with a speed of 400 m/s, passes through a tree its speed is reduced to 200 m/s. How much heat Q is produced and shared by the bullet and the tree? Solution: KEi = KE f + Q where Q is the heat produced. Q = ½ m (vi2 - vf2) = ½ (0.003 kg)[(400 m/s)2 – (200 m/s)2] = 180 J = 180 J x 1 cal / 4.186 J = 43 cal.

7

thermal energy or internal energy: the sum total of all the energy of all the molecules in an object.

temperature (in Kelvins): a measure of the average kinetic energy of individual molecules.

heat: refers to transfer of energy (such as thermal energy) from one object to another due to a difference in temperature.

8

Two equal-mass hot ingots of iron may have the same temperature, but two of them have twice as much thermal energy as one does.

Direction of heat flow depends on temperature. Thus, if 50 g of water at 300C is mixed with 200 g of water at 250C, heat flows from the water at 300C, even though the internal energy of the 250C water is much greater because there is so much more of it.

9

Instruments used to measure temperature are called… thermometers their operation always depends on the expansion of a material (usually mercury or alcohol) with an increase in temperature.

10

Scales used to measure temperature quantitatively are the Celsius (or centigrade) Fahrenheit and Kelvin

11

To convert from Fahrenheit scale to Celsius and vice versa :

0

C = 5/9 ( 0F - 32);

0

F = 9/5 ( 0C )+ 32

Rather than memorize these relations, simply remember that:

0 0C = 50C = 1000C =

320 F 90 F 2120 F 12

Example: Normal body temperature is 98.60 F. What is this in the Celsius scale. Solution:

370 C.

13

The atomic theory of matter states that: If a pure substance were cut into smaller and smaller bits, eventually a smallest piece of that substance would be obtained which could not be divided further. This smallest piece was called an atom, which in Greek means “indivisible.”

14

15

Models of the Atom Experimental data has been the impetus behind the creation and dismissal of physical models of the atom. Rutherford's model, in which electrons move around a tightly packed, positively charged nucleus, successfully explained the results of scattering experiments, but was unable to explain discrete atomic emission—that is, why atoms emit only certain wavelengths of light. Bohr began with Rutherford’s model, but then postulated further that electrons can only move in certain quantized orbits; this model was able to explain certain qualities of discrete emission for hydrogen, but failed completely for other elements. Schrödinger’s model, in which electrons are described not by the paths they take but by the regions where they are most likely to be found, can explain certain qualities of emission spectra for all elements; however, further refinements of the model, made throughout the 20th century, have been needed to explain all observable spectral phenomenon. 16

The law of definite proportion: When two or more elements combine to form a compound, they always do so in the same proportion by weight. E.g., table salt is always formed from 23 parts sodium and 35 parts chlorine; and water is formed from one part hydrogen and eight parts oxygen, by weight. The weight proportions of each element required to form a compound correspond to the relative weights of the combining atoms. By measuring the relative amounts of each element needed to form a large variety of compounds, experimenters established the relative weights of atoms, now referred to as atomic or molecular mass. 17

Unified atomic mass units (u) The relative weights of atoms and molecules, based on assigning the abundant carbon atom, 12C, the value of exactly 12.0000 u The atomic mass of hydrogen, the lightest atom, is 1.0078 u. Values for other atoms are found in the periodic table. The molecular mass of a compound is the sum of atomic masses of the atoms making up the molecule.

18

Brownian movement Named after the biologist Robert Brown, who is credited with its discovery 1827. While observing tiny pollen grains suspended in water, Brown noticed that the tiny grain moved about in tortuous paths, even though the water appeared to be perfectly still. It turned out that Brown’s tiny pollen grains are jostled about by the vigorous barrage of rapidly moving molecules of water.

The atoms (or molecules) of any substance are continually in motion.

19

thermal equilibrium two objects at different temperatures and in thermal contact are said to have reached thermal equilibrium when they eventually get the same temperature.

Zeroth Law of Thermodynamics if two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.

20

Conduction Heat transfer through a material by the collision of atoms.

Convection Heat transfer by a circulating path of fluid particles.

Radiation The process in which energy is transferred by means of

electromagnetic waves.

E.g., visible lights, microwaves, radio waves, radar, and infrared rays.

21

Heat Transfer Heat can be transferred by three processes: conduction, convection, and radiation. Conduction is the transfer of heat along a solid object; it is this process that makes the handle of a poker hot, even if only the tip is in the fireplace. Convection transfers heat through the exchange of hot and cold molecules; this is the process through which water in kettle becomes uniformly hot even though only the bottom of the kettle contacts the flame. Radiation is the transfer of heat via electromagnetic (usually infrared) radiation; this is the principal mechanism through which a fireplace warms a room. Radiation and convection are important to the movement of energy through the Sun. 22

Calculating Heat Transfer specific heat capacity (c) energy needed temperature of 1.0 substance by 1.00C.

the amount of to raise the kg of a

cwater = 4.18 x 103 J/Kg-0C The quantity of heat or gained or lost by a body, Q, is directly proportional to the mass, m, of a the body, its specific heat capacity, c, and the change in the body’s temperature Δt. I.,

Q = mc Δt 23

Sample Problem 1: How much heat is needed to raise the temperature of 2.2 kg of water from 200C to the boiling point? Solution:

m = 2.2 kg c = 4.18 x 103 J/kg-0C Δt = 1000C - 200C = 800C. Q = mc Δt = 2.2 kg x 800C x 4.18 x 103 J/kg-0C = 7.4 J x 105 J Answer

Principle of Heat Exchange When heat is transferred from one body to another, the amount of heat lost by the hot body equals the amount of heat gained by the cold body. Qlost = Qgained m c Δt + m c Δt = 0

24

Sample Problem 2: A 200-g piece of iron (c= 4.5 x 102 J/kg-0C) at 3500C is submerged in 300 g of water at 100C to be cooled quickly. Determine the final temperature of the iron and the water. Solution:

m (iron) = 200 g m (water) = 300 g

Let temperature be tf. Δti = tf - 3500C Δtw = tf - 100C mici Δti + mwcw Δtw = 0 (0.20 kg)(4.5 x 102 J/kg-0C)(tf - 3500C) + (0.30kg)(4.18 x 103 J/kg-0C)(tf 100C) = 0 tf = 330 25

Substance

C (J/kg-0C)

Substance

C (J/kg-0C) 4.18 X 103

Glass

8.4 X 102

Water

Iron

4.5 X 102

Alcohol

2.5 X 103

Brass

3.8 X 102

Ice

2.1 X 103

Silver

2.4 X 102

Steam

2.1 X 103

lead

1.3 X 102

Aluminum

9.2 X 103

26

Solve: •Calculate the amount of heat needed to raise the temperature of the following: (a) 8.4 kg of water by 6.00C (b) 2.1 kg of alcohol by 320C 2. Determine the heat lost when (a) 3.7 kg of water cools from 310C to 240C (b) a 540-g of silver cools from 780C to 140C 3. An electric immersion heater delivers 0.05 MJ of energy to 5.0 kg of a liquid, changing its temperature from 320C to 420C. Find the specific heat capacity of the liquid? 4. Determine how much brass can be heated from 20 to 32 using 1.0 MJ of energy. 27

5. A 2.5-kg pane of glass, initially at 410C, loses 4.2 x 104 J of heat. What is the new temperature of the glass? 6. A 120-g mug at 210C is filled with 210 g of coffee at 910C. Assuming all of the heat lost by the coffee is transferred to the mug, what is the final temperature of the coffee? The specific heat capacity of the mug is 7.8 x 102 J/kg-0C.

28

The First Law of Thermodynamics The increase in internal energy of a system is equal to the heat added plus the work done on the system. Assume that if 10 calories of heat are added to a sample of gas, its temperature rises by 20C. If we add the same 10 calories to a sample of the same gas that has twice the mass, we discover that the temperature rises by only 10C. Adding the same amount of heat does not produce the same rise in temperature because the larger sample of gas has twice as many particles and therefore each particle receives only half as much energy on the average.

29

Absolute Zero Since there is a limit to how much internal energy can be removed from a system, it is reasonable to assume that there is a lowest possible temperature, This is known as absolute zero and has a value of -2730C (0K). The feasibility of reaching this temperature was argued extensively and it was eventually concluded that it was impossible. This belief is formalized in the statement of the

Third Law of Thermodynamics Absolute zero may be approached experimentally but can never reached.

30

Change of State The change from the gaseous state to the liquid state (or from the liquid to the solid), or vice versa, is known as a change of state. While the gas condenses into a liquid, energy continues to leave the system, but the temperature remains the same. Most of this energy comes from the decrease in the electric potential energy between the molecules as they get closer together to form the liquid. The energy that must be released or gained per unit mass of material is known as

latent heat

31

heat of fusion (LF): to the liquid units,

heat required to change 1 kg of a substance from the solid state.

the heat of fusion of water is 79.7 kcal /kg, or, in SI 333 kJ/kg (= 3.33 x 105 J/kg).

heat of vaporization (LV) heat required to change a substance from the liquid to the vapor state. for water, LV is 539 kcal/kg or 2260 kJ/kg. The heats of vaporization and fusion also refer to the amount of heat released by a substance when it changes from a gas to a liquid, or from a liquid to a solid. on the total

The heat involved in a change of phase depends also mass of the substance, i.e.:

Q = mL

32

Change of State

33

Materia Melting l Point (0C) (0C)

Latent Heat (melting) kJ/kg

cal/g

Boiling Point (0C)

Latent Heat (evaporization) kJ/kg

Cal/g

Nitrogen

-210

25.70

6.14

-196

199

47.50

Oxygen

-218

13.80

3.30

-183

213

50.90

0

334.00

79.80

100

2,257

539.00

660

396.00

94.60

2,467

10,900

2,600.0 0

1064

63.00

15.00

2,807

1,710

409.00

Water Aluminu m Gold

34

35

How much energy is released when 5 kg of water freezes at 00C? Answer: Q = mL = (5.00 kg)(3.33 X 105 J/kg) = 1.67 x 106J of energy released. Examples:

1. Making ice. How much energy does a refrigerator have to remove from 1.5 kg of water at 200C to make ice at – 120C? Solution:

Heat must flow out to reduce the water from 200C to 00C, to change it to ice, then to lower the ice from 00C to – 120C,

Q = mcw (200C - 00C) + mLF + mcice[00C – (- 120C)] = 1.5 kg(4186 J/kg-0C)(200C) +1.5 kg(3.33 x105 J/kg) + 1.5 kg(2100 J/kg- 0C)(120C) = 6.6 x 105 J = 660 kJ 36

2. Will all the ice melt? At a party, a 0.50-kg chunk of ice at – 100C is placed in 3.0 kg of “iced” tea at 200C. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Solution:

Before we can write down an equation, we must first check to see if the final state will be all ice, a mixture of ice and water at 00C, or all water. To bring the 3.0 kg of water at 200C down to 00C would require an energy release of

mwcw (200C - 00C) = 3.0 kg(4186 J/kg-0C)(200C) = 250 kJ On the other hand, to raise the ice from - 100C to 00C, would require micecice [00C – (- 100C)] = 0.50 kg(2100 J/kg-0C)(100C) = 10.5 kJ and to change the ice to water at 00C would require miceLF = 0.50 kg(333 kJ/kg) = 167 kJ 37

for a total of 10.0 kJ + 167 kJ = 177 kJ. This is not enough energy to bring the 3.0 kg of water at 200C down to 00C, so we know that the mixture must end up all water, somewhere between 00C and 200C. Now we can determine the final temperature T by applying the conservation of energy and writing heat to raise 0.50 kg of ice from - 100C to 00C, plus heat to change 0.50 kg of ice to water, plus + heat to raise 0.50 kg of water from 00C to T equals heat lost by 3.0 kg of water cooling from 200C to T, so 10.5 kJ + 167 kJ + 0.50 kg(4186 J/kg-0C)(T) = 3.0 kg(4186 J/kg-0C)(200C – T) 14,000T = 73,800 T = 5.10C

38

39

Condensation Condensation, in physics, is the process of reduction of matter into a denser form, as in the liquefaction of vapor or steam. Condensation is the result of the reduction of temperature by the removal of latent heat of evaporation, the liquid product being known as condensate. The removal of heat shrinks the volume of the vapor and decreases the velocity of, and the distance between, molecules. According to the kinetic theory of behavior of matter, the loss of energy will lead to the transformation of the gas into a liquid condensate. Condensation is an important part of the process of distillation and in the operation of steam engines, where exhaust steam is converted back into water by a device called a condenser. In meteorology, both the formation of clouds and the precipitation of dew, rain, and snow are known as condensation. In chemistry, condensation is a reaction involving the union of atoms in the same or different molecules. The process often leads to the elimination of a simple molecule such as water or alcohol to form a new and more complex compound, often of greater molecular weight.

40

Thermal Expansion When the temperature increases, nearly all materials expand, but not all materials expand at the same rate. Solids, being most tightly bound, expand the least, while all gases expand at the same rate. Each material’s characteristic thermal expansion is reflected in a number called its coefficient of expansion. The coefficient of expansion gives the frictional change in the size of the object per degree change in temperature.

41

The expansion for a particular object is given by:

ΔL = αLΔT where ΔL = change in length α = coefficient of expansion ΔT = change in temperature L = original length 42

…Thermal Expansion Over small temperature ranges, the linear nature of thermal expansion leads to expansion relationships for length, area, and volume in terms of the linear expansion coefficient .

43

Linear Expansion The relationship governing the linear expansion of a long thin rod can be reasoned out as follows:

44

One notable exception to this general rule is water. It usually undergoes a normal thermal expansion, but as it approaches its freezing point, it will expand instead of shrink. As an object expands or contracts with a temperature change, its change in length depends on three quantities: the original length, the temperature change, and the thermal properties of the material composing the object. 45

Think about a metal rod that is 3 ft (1 m) long and has its temperature increased to the point where it expands by one hundredth of a millimeter. Now consider two such identical rods placed end to end. They are the equivalent of a single rod 6 ft (2 m) long that will expand by two hundredths of a millimeter for the same temperature increase. Hence a larger object will have a greater change in length with temperature than a smaller object, simply because the larger object has more material to expand.

46

Secondly, the change in length also depends on the temperature change. An object will expand twice as much for a 20° temperature change as for a 10° temperature change. In a third variable, different materials also expand at different rates. For example, with a given temperature change, a block of wood will not expand as much as a similar block of metal. The coefficient of thermal expansion, found by experimentation, is a property of the material that accounts for the different expansion rates for different materials. Using these three factors, the change in length of an object is equal to the coefficient of thermal expansion multiplied by the original length multiplied by the temperature change. 47

Expansion Concepts

48

49

Thermal Expansion Coefficients at 20 C Material

Fractional expansion per degree C x10^-6

Fractional expansion per degree F x10^-6

Glass, ordinary

9

5

Glass, pyrex

4

2.2

Quartz, fused

0.59

0.33

Aluminum

24

13

Brass

19

11

Copper

17

9.4

Iron

12

6.7

Steel

13

7.2

Platinum

9

5

Tungsten

4.3

2.4

Gold

14

7.8

Silver

18

10 50

Problem: By how long does a 50-meter steel bridge expand overnight if its temperature increases by 400C overnight? The coefficient of expansion of steel is 0.000011 meter for each meter of length for each degree Celsius in temperature. Solution: ΔL = αLΔT = 0.000011/ 0C x (50 m)(400C) = 0.022 m

51

Conceptual Questions •A block of wood falls from a chair and lands on the floor. What happens to the kinetic energy of the block? The ke is converted to thermal energy (and energy of distortion) 2. What does the zeroth law of thermodynamics tell us about measuring the temperature of an object? Two objects in thermal equilibrium have the same temperature. 3. Can two objects be in thermal equilibrium when they are not touching? Explain. Yes, if they have the same temperature. 4. How do units of heat and temperature differ? The units do not represent the same physical quantity (Joule & Kelvin) 52

5. What is the difference between heat and temperature? Heat is a flow of thermal energy, whereas temperature is the average kinetic energy of the atoms and molecules. 6. How do the internal energies of a cup of water and a gallon of water at the same temperature compare? The gallon of water has much more internal energy. 7. Under what condition is the first law of thermodynamics valid? Under all conditions. 8. Does it take more thermal energy to raise the temperature of 5 grams of water than 5 grams of ice by 6C? The water because it has a larger specific heat.

53

9. How is specific heat defined?

The heat added divided by product of the mass and the resulting temperature change. 10.Why do climates near the coasts tend to be more moderate than in the middle of the continent? It takes tremendous amount of thermal energy to change the temperature of the water in the ocean. 11. Given that the melting point and freezing temperatures of water are identical, which determines whether a mixture of ice and water will freeze or melt? It will freeze if heat is removed, melt if heat is added. 12. Why can an iceberg survive floating for several weeks in seawater that’s above freezing? 54

It requires a lot of thermal energy to melt the ice without changing its temperature. Also, ice is not a very good conductor of thermal energy 13. What would happen to a pot of water on the stove if there were no latent heat of vaporization required for converting water to steam? The water would vaporize instantly when it reached the boiling temperature. 14. If you put an unwrapped steak in the freezer, it freezes and then overtime dries out. Why? The ice in the steak sublimes. 15. An old biker’s adage is, “If your feet get cold, put on a hat.” What is the physics behind this? A lot of thermal energy is lost from the head.

55

16. Why is it possible to hold a lighted match until the flame burns quite close to your finger? Wood is a poor thermal conductor.

56

Heat Engines and the Second Law of Thermodynamics A "heat engine" is a machine which converts internal energy (from a high temperature body) into some other form of energy. When the lid on a pan of boiling water is lifted by the steam inside, the internal energy of the steam is being converted to gravitational potential energy of the lid. This is a very simple example of a heat engine. The conversion of energy from some other form of energy to internal energy of a substance can be done with 100% efficiency of conversion. For example, 100J of electrical energy will be converted to 100J of internal energy by a resistor. Conversion of energy from internal energy to some other form cannot be done with the same efficiency. 57

To show this, consider the possible heat engine shown below. (The diagrams are very simplified but are similar to some real heat engines).

Gas expanding, doing work, pushing piston down.

To obtain more work, we must now push the piston back up 58

If we simply push the piston back up we will do just as much work as we obtained during the expansion stage: not a very satisfactory situation! To make sure that we do less work pushing the piston back to its original position we could allow the gas to cool down before pushing the piston up. Or we could let this still quite hot gas escape, push the piston up and then put in fresh gas. However you do it you will always find that you need to allow the gas to give up some of its internal energy to the surroundings (as it cools down). Therefore, the conversion of energy from internal energy of the hot gas can never be 100% efficient.

59

This is one illustration of the second law of thermodynamics which can be stated in many different ways. One statement is as follows

No heat engine, operating in a continuous cycle, can do work without transferring some internal energy from a hot body to a cold body.

That is, even the best theoretical heat engine cannot convert 100% of the incoming heat into work. The hot body is called the heat source and the cold body (often the surroundings) is called the heat sink

60

Question: How much work does a heat engine perform if it extracts 100 joules of energy from a hot region and exhausts 60 joules to a cold region? Answer: Conservation of energy requires that the work be equal to 40 joules, the difference between the input and the output. Real Engines In the case of heat engines the efficiency is ɳ of an engine is equal to the ratio of the work W produced divided by the heat Q from the hot region.

ɳ = W/Q

61

Carnot’s ideal heat engine has the maximum theoretical efficiency which can be expressed as a simple relationship using Kelvin temperatures:

ɳ = 1 – Tc/Th Carnot’s efficiency is larger if the temperature of the exhaust region Tc is low or that of the hot region is high. Given that Carnot’s engines cannot run much hotter than 5500C, and that temperatures cannot be much lower than 500C, calculate the maximum theoretical efficiency of the steam engines. ɳ = 1 – Tc/Th = 1 – 323 K / 823 K = 0.61 = 60% 62

Question: What is the maximum theoretical efficiency for a heat engine running between 1270C and 270C? Answer:

ɳ = 1 – 300 K/400 K = 1 – 0.75 = 25%

63