Chapter 8: Temperature and Heat s1

MR. SURRETTE VAN NUYS HIGH SCHOOL

CHAPTER 8: TEMPERATURE AND HEAT TEMPERATURE AND GAS LAWS CLASS NOTES

THERMAL PHYSICS Thermal physics describes the effects of kinetic energy on the molecular scale.

TEMPERATURE Temperature is a measurement of the average kinetic energy of molecules:

3 KAVG = /2 kBT -23 (kB = Boltzmann’s constant = 1.38x10 J/K)

TEMPERATURE SCALES 1. The degree Fahrenheit (oF) non-metric temperature scale was devised so that the freezing and boiling temperatures of water are whole numbers. 2. The degree Celsius (oC) scale was devised by dividing the range of temperature between the freezing and boiling temperatures of pure water into 100 equal parts. 3. The Kelvin (K) temperature scale is an extension of the degrees Celsius scale down to absolute zero, a hypothetical temperature characterized by a complete absence of heat energy.

TEMPERATURE SCALE COMPARISONS

TEMPERATURE CONVERSION EQUATIONS TC is the Celsius temperature and T is the Kelvin temperature. The size of a degree on the Kelvin scale is identical to the size of a degree on the Celsius scale.

TEMPERATURE CONVERSION EQUATIONS Some temperature conversions are: TC = T – 273 9 TF = ( /5)TC + 32 5 TC = /9 (TF – 32)

1 | P a g e PHYSICS MR. SURRETTE VAN NUYS HIGH SCHOOL

Example 1. Oxygen condenses into a liquid at approximately 90o Kelvin (90 K). What temperature is this in degrees Fahrenheit? 1A. 9 (1) TF = /5TC + 32 9 (2) TF = /5(T – 273) + 32 9 (3) TF = /5(90 – 273) + 32 o (4) TF = - 297.4

LINEAR EXPANSION The following are two equations for the thermal expansion of a solid. The constant  called the average temperature coefficient of linear expansion (note: this  has nothing to do with , the angular rate of acceleration).

LINEAR EXPANSION EQUATION

L = L0T (L = change in length [m],  = coefficient of linear expansion [1/oC], LO = original length [m], T = change in temperature [oC])

LINEAR EXPANSION EQUATION

L – L0 = L0(T – T0) (L = final length [m], T = final temperature [oC])

YOUNG’S MODULUS Young’s modulus is a measure of the resistance of a body to elongation. It is sometimes used to compute linear expansion problems.

YOUNG'S MODULUS EQUATION

Y = (F/A) / (L/L0) (Y = Young’s modulus [N/m2], F = force [N], A = cross-sectional area [m2], L = change in length [m], LO = original length [m])

Example 2. Suppose the ends of a 50-m-long steel rail are rigidly clamped at - 5o C to prevent expansion. The rail has a cross-sectional area of 37 cm2. What force does the rail exert when it is o -5 o 11 2 heated to 42 C? (STEEL = 1.1 x 10 / C, YSTEEL = 2 x 10 N/m ) 2A. (1) Y = (F/A) / (L/L) (2) (L/L) = (F/A) / Y (3) Y(L/L) = (F/A) (4) (F/A) = Y(L/L) (5) F = (AYL) / L (6)  L = L  T: F = (AYLT) / L (7) F = AYT (8) F = (3.7 x 10-3 m2)(2 x 1011 N/m2)(1.1 x 10-5 /oC)(47oC) (9) F = 3.83 x 105 N

AREA AND VOLUME EXPANSION The quantity  (gamma) is the average temperature coefficient of area expansion and  (beta) is the average temperature coefficient of volume expansion.  = 2 and  = 3:

A = A0T

V = V0T

Example 3. Suppose one sphere is made of a metal that has twice the coefficient of linear expansion of a second sphere. The coefficient of volume expansion for the first sphere will be bigger than the second by a factor of what number? 3A. Since  for both spheres = 3, the first sphere will expand twice as much.

THE KINETIC THEORY OF GASES The model of an IDEAL GAS is based on the following assumptions: 1. The number of molecules is large, and the average separation between them is large. 2. The molecules obey Newton’s laws of motion, but the individual molecules move in a random fashion. 3. The molecules undergo elastic collisions with each other. 4. The forces between molecules are negligible, except during a collision. 5. The gas under consideration is a pure gas. 6. The gas is in thermal equilibrium with the walls of the container.

AVOGADRO’S NUMBER Equal volumes of gas at the same temperature and pressure (STP conditions) contain the same number of molecules. Single moles of any gas at standard temperature and pressure contain the same number of molecules: 23 NA = 6.02 x 10 / mole

STANDARD TEMPERATURE AND PRESSURE The volume of a gas is usually measured at standard temperature and pressure (STP). Standard temperature is 0o C. Standard pressure is 1 atmosphere (atm). At STP, 1 mole of any gas occupies a volume of 22.4 L.

IDEAL GAS EQUATIONS In the following ideal gas equation, T is the temperature in kelvins and n is the number of moles. R is the universal gas constant and p is pressure: pV = nRT (R = 8.31 J/mol.K or 0.0821 L.atm/mol.K)

Example 4. Three moles of an ideal gas are confined to a 30 liter container at a pressure of 2 atmospheres. What is the gas temperature? (R = 0.0821 L.atm/mole.K) 4A. (1) pV = nRT (2) T = (pV) / (nR) (3) T = [(2 atm)(30 liters)] / [(3 mol)(0.0821 L.atm/mol.K)] (4) T = 243.6 K

IDEAL GAS EQUATIONS Let p1, V1, and T1 be initial conditions. Let p2, V2, and T2 be final conditions. T, the absolute temperature, is in kelvins. Therefore:

(p1V1) / T1 = (p2V2) / T2

ROOT MEAN SQUARE The expression for the root mean square (rms) speed shows that, at a given temperature, lighter molecules move faster on the average than heavier molecules (M = molecular mass). In particular: 1/2 VRMS = (3RT / M) 1/2 VRMS = (3kBT / ) -23 (kB = Boltzmann’s constant = 1.38 x 10 J/K)  = mass of one molecule)

Example 5. If the rms velocity of a helium atom at room temperature is 1350 m/s, what is the rms velocity of an oxygen (O2) molecule (mass of O2 = 32 grams, mass of He = 4 grams)? 5A. 1/2 (1) vRMS = [(3RT)/M] 1/2 (2) vRMS O2 is proportional to (1/32 grams) 1/2 (3) vRMS He is proportional to (1/4 grams) 1/2 1/2 (4) vRMS (O2 / He) is proportional to (32) / (4) (5) vRMS (O2 / He) = 2.8 (6) vRMS = 1350 m/s / 2.8 (7) vRMS = 477.3 m/s

CHAPTER 8: TEMPERATURE AND HEAT HEAT CLASS NOTES

HEAT VERSUS TEMPERATURE When two systems at different temperatures are in contact, energy will transfer between them until they reach the same temperature. This energy is called heat, or thermal energy, Q.

TEMPERATURE CHANGES Two objects are in thermal contact if they can affect each other’s temperature. Thermal equilibrium exists when two objects in thermal contact no longer affect each other’s temperature.

THERMAL CONTACT VERSUS EQUILIBRIUM If a carton of milk from the refrigerator is set on the kitchen counter top, the two objects are in thermal contact. After several hours, their temperatures are the same, and they are then in thermal equilibrium.

UNITS OF HEAT One unit of heat is the calorie, defined as the amount of heat needed to raise the temperature of one gram of water 1oC. Another unit of heat is the joule: 1 calorie = 4.186 joules

SPECIFIC HEAT The amount of heat needed to increase the temperature of a given substance varies. Every substance has a unique specific heat value, c: Q = mcT Q = heat [joules or calories] m = mass [grams or kilograms] c = specific heat [joules/kg.oC] T = change in temperature [oC]

LATENT HEAT The heat needed to cause a phase change depends on the value of latent heat. The latent heat of fusion, LF, is used when going from solid to liquid and the latent heat of vaporization, LV, is used when going from liquid to gas.

LATENT HEAT EQUATION Q = mL Q = heat [joules or calories] m = mass [grams or kilograms] L = latent heat [joules/kg]

TOTAL HEAT ENERGY Total heat energy is equal to the sum of specific heat and latent heat:

Q = QC + QL

Example 1. How much heat energy is required to melt a 20 gram block of silver at 20oC? The melting point of silver is 960o C, its specific heat is 2.34 x 102 J/kg.oC, and its heat of fusion is 8.83 x 104 J/kg. 1A.

(1) QC = mcT (2) (0.020 kg)(2.34 x 102 J/kg.oC)(960oC – 20oC) 3 (3) QC = 4.40 x 10 J (4) QL = mL 4 (5) QL = (0.020 kg)(8.83 x 10 J/kg) 3 (6) QL = 1.77 x 10 J (7) Q = QC + QL (8) Q = (4.40 x 103 J) + (1.77 x 103 J) (9) Q = 6.17 x 103 J

Example 2. A 3 gram copper penny at 20oC is dropped from a height of 300 m and strikes the ground. If 60% of the energy goes into increasing the internal energy of the coin, determine its final temperature. (The specific heat of copper is 387 J/kg.oC) 2A. (1) Ug = mgh 2 (2) Ug = (0.003 kg)(9.8 m/s )(300 m) (3) Ug = 8.82 J (4) 60% Ug = 5.29 J (5) Q = mcT (6) T = Q / mc (7) T = 5.29 J / (0.003 kg)(387 J.kg.oC) (8) T = 4.56oC (9) T + T = 20oC + 4.56oC = 24.6oC

CONSERVATION OF HEAT ENERGY The heat given off by a substance in a closed environment must be equal to the heat absorbed by its environment: QOUT = QIN

Example 3. A 1 kg block of copper at 20oC is dropped into a large vessel of liquid nitrogen at 77 K. How many kilograms of nitrogen boil away by the point the copper reaches 77 K? (The specific heat of copper is 385 J/kg.oC. The heat of vaporization of nitrogen is 2 x 105 J/kg.) 3A. (1) Tc = T – 273 (2) T = Tc + 273 (3) T = 20 + 273 (4) T = 293 K

(5) QC = mcT o (6) QC = (1 kg)(385 J/kg. C)(293 K – 77 K) 4 (7) QC = 8.32 x 10 J (8) QOUT = QIN (9) QC = QL (10) QL = mL

3A. (continued...) (11) m = QL / L (12) m = 8.32 x 104 J / 2 x 105 J/kg (13) m = 0.42 kg

Example 4. A solar heating system has a 25% conversion efficiency; the solar radiant incident on the panels is 500 W/m2. What is the increase in temperature of 30 kg of water in a one hour period by a 4.0 m2 area collector? (The specific heat of water is 4186 J/kg.oC) 4A. (1) Total radiation: 4.0 m2 (500 W / 1.0 m2) = 2000 W (2) Effective radiation: 2000 W x 25% = 500 W (3) 500 W = 500 Joules / second (4) Determine Q: 3600 s (500 J / s) = 1.8 x 106 J (5) Q = mcT (6) T = Q / mc (7) T = 1.8 x 106 J / (30 kg)(4186 J/kg.oC) (8) T = 14.3oC

TYPES OF HEAT TRANSFER There are three basic processes of heat transfer: Conduction Convection Radiation

CONDUCTION Conduction transfers heat when there is a temperature gradient across a body. For example, if you heat a metal rod at one end with a flame, heat will flow from the hot end to the colder end.

LAW OF HEAT CONDUCTION The constant k represents thermal conductivity. The rate of heat energy that flows is the heat current, H. Conduction only occurs when there is a temperature difference.

LAW OF HEAT CONDUCTION EQUATION H = kAT) / d H = heat current [W] k = thermal conductivity [W/moC] A = cross-sectional area [m2] T = change in temperature [oC] d = distance across conductor [m]

Example 5. A silver bar of length 30 cm and cross-sectional area 1 cm2 is used to transfer heat from a o o o 100 C reservoir to a 0 C reservoir. How much heat is transferred per second? (kSILVER = 427 W/m C) 5A. (1) H = kA(T) / d (2) (427 W/moC)(1 x 10-4 m2)(100oC) / (0.3 m) (3) H = 14.23 W

CONVECTION When heat transfer occurs as the result of the motion of material, such as the mixing of hot and cold fluids, the process is referred to as convection.

RADIATION Heat transfer by radiation is the result of the continuous emission of electromagnetic radiation by all bodies.

STEFAN’S LAW The rate of emission of heat energy by a black-body (star) is given by Stefan’s Law:

P = AeT4  = Stefan-Boltzmann constant = 5.6696 x 10-8 W/m2, A = surface area [m2], T = absolute temperature [K], e (emissivity) = 0 – 1)

Example 6. If the absolute surface temperature of a spherical object were tripled, by what factor would the rate of radiated energy emitted from its surface be changed? 6A. (1) P = AeT4 (2) P = Ae(3T)4 (3) P = 81AeT4 (4) P = 81 times

CHAPTER 8: TEMPERATURE AND HEAT THERMODYNAMICS CLASS NOTES

FIRST LAW OF THERMODYNAMICS An increase in one form of energy must be accompanied by a decrease in some other form of energy: U = Q – W

FIRST LAW OF THERMODYNAMICS U is change in internal energy (the random motion of molecules), Q is the heat added to the system, and W is the work done by the system. [Q is positive when heat enters the system and negative when heat leaves the system.]

Example 1. An ideal gas undergoes an adiabatic process (no heat exchanged) while it performs 25 J of work on its environment. What is its change in internal energy? 1A. (1) U = Q - W (2) U = 0 – 25 J (3) U = - 25 J

WORK AND HEAT The work done by an expanding gas is the area under the curve in a PV (Pressure/Volume) diagram.

WORK AND HEAT

WORK AND HEAT EQUATION

W = pV Work [J] = Pressure [Pa = N/m2] x Volume [m3]

IDEAL GAS EQUATIONS Let p1, V1, and T1 be initial conditions. Let p2, V2, and T2 be final conditions. T, the absolute temperature, is in kelvins. Therefore: (p1V1) / T1 = (p2V2) / T2 [p = pressure, V = volume, T = temperature]

Example 2. A bottle containing an ideal gas has a volume of 2 m3 and a pressure of 1 x 105 N/m2 at a temperature of 300 K. The bottle is placed against a metal block that is maintained at 900 K and the gas expands as the pressure remains constant until the temperature of the gas reaches 900 K. How much work is done by the gas? 2A. (1) (p1V1) / T1 = (p2V2) / T2 (2) V2 = (p1V1T2) / (T1p2) 5 3 5 (3) V2 = [(1 x 10 Pa)(2 m )(900 K)] / [(300 K)(1 x 10 Pa)] 3 (4) V2 = 6 m

(5) V = V2 – V1 (6) V = 6 m3 – 2 m3 = 4 m3 (7) W = pV (8) W = (1 x 105 Pa)(4 m3) (9) W = 4 x 105 J

THERMODYNAMIC PROCESSES Heat performs work under various conditions. These conditions are called thermodynamic processes. 1. A cyclic process originates and ends at the same state. The work done per cycle equals the heat added. 2. An adiabatic process is one in which no heat enters or leaves the system; that is, Q = 0. 3. An isobaric process occurs at constant pressure. 4. An isovolumetric process occurs at constant volume. W = 0 for such a process: W = pV = p(0) = 0.

HEAT ENGINES A heat engine is a device that converts thermal energy to other useful forms, such as electrical and mechanical energy. The engine absorbs a quantity of heat, QH, from a hot reservoir, does work W, and then gives up heat QC, to a cold reservoir.

EFFICIENCY The thermal efficiency, e, of a heat engine is the ratio of the net work done to the heat absorbed: e = (QH – QC) / QH e = W / QH

Example 3. A gasoline engine absorbs 2500 J of heat energy and performs 500 J of mechanical work in each cycle. What is the efficiency of the engine? 3A. (1) e = (QH – QC) / QH (2) QH – QC = Work (3) e = Work / QH (4) e = 500 J / 2500 J (5) e = 20%

THE CARNOT CYCLE The most efficient cyclic process is called the Carnot cycle.

THE CARNOT CYCLE

CARNOT ENGINE EFFICIENCY The thermal efficiency of the ideal Carnot engine depends on the temperatures of the hot and cold reservoirs: ec = (TH – TC) / TH

ENTROPY Entropy is a quantity used to measure the disorder of a system. For example, the molecules of a gas in a container at a high temperature are in a more disorganized state (higher entropy) than the same molecules at a lower temperature:

ENTROPY EQUATION S = Q / T S (entropy) is the ratio of heat added to a system to the absolute temperature.

SECOND LAW OF THERMODYNAMICS The entropy of an isolated system always increases if the system undergoes an irreversible process. If an isolated system undergoes a reversible process, the entropy remains constant. The entropy of the universe increases in all natural processes.

Example 4. Calculate the entropy change when 1 mol (18 grams) of water at 100oC is converted into steam (The heat of vaporization of water is 540 cal/g). 4A. (1) Q = mL (2) Q = (18 g)(540 cal/g) (3) Q = 9720 cal (4) 9720 cal (4.186 J/cal) = 40,688 J (5) S = Q / T (6) S = 40,688 J / 373 K (7) S = 109.1 J/K

11 | P a g e PHYSICS