Chapter 17 Temperature and Heat PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman Lectures by James Pazun Modified by P. Lam 6_17_2012
Topics for Chapter 17 Condition for thermal equilibrium. temperature scales thermal expansion thermal stress heat, phase changes, and calorimetry heat flows with convection, conduction, and radiation (qualitative)
Condition for thermal equilibrium A hot cold B A B Non-equilibrium T A > T B Net energy flows from A to B Equilibrium T A = T B No net energy flow T A =temperature of system A T B =temperature of system B
The zero th law of thermodynamics If A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A is also in thermal equilibrium with B Simply stated: If T A =T C and T B =T C, then T A =T B
Application of the zero th law When the doctor takes your temperature, the doctor implicitly invokes the zero th law because the doctor assumes the thermometer has reached thermal equilibrium with your body and reading the temperature of the thermometer is the same as reading your temperature.
Relating the three popular temperature scales Values on the temperatures scales (Fahrenheit, Centigrade/Celsius, and Kelvin) may be readily interconverted. See Figure below. o C + 273! K & 180 ( o C) + 32 = o F 100 Absolute zero = absolute lowest temperature.
Calibration of Thermometers There are many physical properties which vary with temperature and hence can be used as a thermometer. Here are two examples: (1) Expansion and contraction of a liquid such as mercury or alcohol as the temperature increases or decreases, respectively. (2) Pressure of a gas inside a constant volume container increases as temperature rises. Pressure " T (in Kelvin) Thermometers are calibrate with some reproducible temperatures such as (i) (ii) Melting point of ice at 1 atmospheric pressure Boiling point of water at 1 atmospheric pressure
Thermal expansion linear and volume expansion A change in length accompanies a change in temperature. The amount of the change depends on the material.!l "!!T L o! = linear expansion coefficient of the material unit of! = 1/Kelvin Similarly for volume expansion:!v " "!T; V o for "small"!t, " # 3!
Coefficients of expansion
Thermal stress Thermal expansion joints allow roads to expand and contract freely without causing stress to the materials. Thermal stress - Stress (force/ area) develops in a material if it is not allowed to expand or contract as temperature changes. F A = Y "L = Y# "T ; Y = Young's modulus of the material L o The textbook use a negative sign to remind us the direction of the force F A = $Y#"T I think that it is not necessary and may be even confusing (one more convention to remember!) You can figure out which direction is the force using common sense.
Thermal stress Example 17.5: An aluminum cylinder (10 cm long with crosssectional area =20cm 2 ) is used at a spacer between two rigid walls. At T=17.2 o C it just fit between the two walls. When it warms up to 22.3 o C, what is the stress (force/ area) in the cylinder and the total force it exerts on each wall? Given the Young s modulus for aluminum is 7x10 10 N/m 2 and is linear expansion coefficient is 2.4 x10-5 /K
Specific heat and heat capacity The specific heat of a substance reveals how much temperature will change when a given amount of heat is added or removed from the substance. Q = mc!t m = mass; c = specific heat [J/kg K] mc= heat capacity [J/K] Water is a benchmark as one gram of water will absorb 1 cal of heat to raise its temperature by 1 o C. c =1 cal g C o = 4190 J kg K
Specific heat values Note:Specific heat varies with temperature and pressure. These values are valid for a limited range of temperature and pressure only.
Phase changes and temperature behavior Example: Start with 1 kg of ice at -20 o C and add heat. Pressure=1 atm. T Not drawn to scale! Heat of vaporization gas 100 o C 0 o C -20 o C solid Heat of fusion Liquid Heat input (kj) 42kJ (42+334)=376KJ (376+419)=795kJ (795+2256)kJ
Heats of Fusion and Heats of Vaporization
Calorimetry Example 17.8: (a) A 0.5 kg aluminum cup is initially at T=150 o C. 0.3kg of water at T=70 o C is poured into the cup. Assume no heat exchange with the surrounding, find the final temperature of the cup and the water. "Q total = 0 = m Al c Al (T f # T i Al ) + m w c w (T f # T i w ) 0 = (0.5)(910)(T f #150) + (0.3)(4190)(T f # 70) $ T f = 91.3 o C (b) What happens if the initial temperature of the aluminum cup were 200 o C? "Q total = 0 = m Al c Al (T f # T i Al ) + m w c w (T f # T i w ) 0 = (0.5)(910)(T f # 200) + (0.3)(4190)(T f # 70) $ T f =104.6 o C!! What does it mean?
Calorimetry Example 17.9 (change in temperature and phase): 0.25 kg of water in a cup is initially at T=25 o C. An amount ice (initially at T= - 20 o C) is poured into the cup. How much ice is needed so that the final temperature will be O o C and no ice left. Assume no heat exchange with the surrounding and the cup has negligible heat capacity.
Methods of heat transfer (qualitative) Heat conduction through a solid. Heat current is (1) proportional to the temperature difference (2) proportional to the crosssectional area (A) (3) inversely proportional to the length of the solid Heat current ( Joule s = Watt) " dq dt = ka T # T H C L $ W ' % & m K( ) k = thermal conductivity of the solid
Convection of heat (qualitative) Heat transfer in a fluid (gas or liquid). Heat energy is carried by the movement of the hot fluid particles (if movement is restricted than there is no convection) Hot fluid (less dense) rises while cold fluid (more dense) decends. Gravity is necessary for convection.
Radiation of heat (qualitative) A hot object can lose energy by emitting electromagnetic radiations. On the right, an infrared photograph shows the infrared radiation given off by a person. hot cold Radiation heat current (Watt) = Ae"T 4 T = temperature of the hot object in Kelvin! A = cross - sectional area e = emissivity (depends on the nature of the object) (0 < e < 1) " = a fundamental constant # 5.67x10-8 W m 2 K 4