# Ch20 ssm

18. Dec 2018
1 von 9

### Ch20 ssm

• 1. Chapter 20 Thermal Properties and Processes Conceptual Problems 3 • Why is it a bad idea to place a tightly sealed glass bottle that is completely full of water, into your kitchen freezer in order to make ice? Determine the Concept Water expands greatly as it freezes. If a sealed glass bottle full of water is placed in a freezer, as the water freezes there will be no room for the expansion to take place. The bottle will be broken. 9 •• The phase diagram in Figure 20-15 can be interpreted to yield information on how the boiling and melting points of water change with altitude. (a) Explain how this information can be obtained. (b) How might this information affect cooking procedures in the mountains? Determine the Concept (a) With increasing altitude, decreases; from curve OC, the temperature of the liquid-gas interface decreases as the pressure decreases, so the boiling temperature decreases. Likewise, from curve OB, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased. 13 •• Two solid cylinders made of materials A and B have the same lengths; their diameters are related by dA = 2dB. When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations? (a) kA = kB/4, (b) kA = kB/2, (c) kA = kB, (d) kA = 2kB, (e) kA = 4kB Picture the Problem The rate at which heat is conducted through a cylinder is given by xTkAdtdQI ΔΔ== // (see Equation 20-7) where A is the cross- sectional area of the cylinder. The heat current in cylinder A is the same as the heat current in cylinder B: BA II = Substituting for the heat currents yields: L T Ak L T Ak Δ = Δ BBAA ⇒ A B BA A A kk = 409
• 2. Chapter 20410 Because dA = 2dB: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = B B BA 4A A kk ⇒ B4 1 A kk = )(a is correct. Estimation and Approximation 17 •• You are using a cooking pot to boil water for a pasta dish. The recipe calls for at least 4.0 L of water to be used. You fill the pot with 4.0 L of room temperature water and note that this amount of water filled the pot to the brim. Knowing some physics, you are counting on the volume expansion of the steel pot to keep all of the water in the pot while the water is heated to a boil. Is your assumption correct? Explain. If your assumption is not correct, how much water runs over the sides of the pot due to the thermal expansion of the water? Determine the Concept The volume of water overflowing is the difference between the change in volume of the water and the change in volume of the pot. See Table 20-1 for the coefficient of volume expansion of water and the coefficient of linear expansion of steel. Express the volume of water that overflows when the pot and the water are heated: ( ) TV TVTV VVV Δ ΔΔ ΔΔ 0steelOH 0steel0OH potOHovefrlow 2 2 2 ββ ββ −= −= −= Because the coefficient of volume expansion of steel is three times its coefficient of linear expansion: steelsteel 3αβ = Substituting for andOH2 β steelβ yields: ( ) TVV Δ3 0steelOHoverflow 2 αα −= Substitute numerical values and evaluate :overflowV ( )( )( )( ) mL56C20C100L4.0K10113K10207.0 1613 overflow =°−°×−×= −−−− V Your assumption was not correct and 56 mL of water overflowed.
• 3. Thermal Properties and Processes 411 19 •• Estimate the thermal conductivity of human skin. Picture the Problem We can use the thermal current equation for the thermal conductivity of the skin. If we model a human body as a rectangular parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface area is about 1.8 m2 . We’ll also assume that a typical human, while resting, produces energy at the rate of 120 W, that normal internal and external temperatures are 33°C and 37°C, respectively, and that an average skin thickness is 1.0 mm. Use the thermal current equation to express the rate of conduction of thermal energy: I = kA ΔT Δx ⇒ x T A I k Δ Δ = Substitute numerical values and evaluate k: ( ) Km mW 17 m100.1 C33C37 m8.1 W120 3 2 ⋅ = × °−° = − k 25 ••• You are in charge of transporting a liver from New York, New York to Los Angeles, California for a transplant surgery. The liver is kept cold in a Styrofoam ice chest initially filled with 1.0 kg of ice. It is crucial that the liver temperature is never warmer than 5.0°C. Assuming the trip from the hospital in New York to the hospital in Los Angeles takes 7.0 h, estimate the R-value the Styrofoam walls of the ice chest must have. Picture the Problem The R factor is the thermal resistance per unit area of a slab of material. We can use the thermal current equation to express the thermal resistance of the styrofoam in terms of the maximum amount of heat that can enter the chest in 7.0 h without raising the temperature above 5.0°C. We’ll assume that the surface area of the ice chest is 1.0 m2 and that the ambient temperature is 25°C The R-factor needed for the styrofoam walls of the ice chest is the product of their thermal resistance and area: RAR =f (1) Use the thermal current equation to express R: tottot ΔΔ Δ ΔΔ Q tT t Q T I T R === Substitute for R in equation (1) to obtain: tot f ΔΔ Q tTA R =
• 4. Chapter 20412 The total heat entering the chest in 7 h is given by: OHOHicefice waterice warm ice melttot 22 ΔTcmLm QQQ += += Substitute for Qtot and simplify to obtain: ( )OHOHfice f 22 Δ ΔΔ TcLm tTA R + = Substitute numerical values and evaluate Rf: ( ) ( ) ( ) Btu fthF 8 C5 Kkg kJ 18.4 kg kJ 5.333kg1 Btu J35.1054 h7 C5 F9 C20 m1029.9 ft1 m1.0 222 2 2 f ⋅⋅° ≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ °⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⋅ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ° ×°⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × = − R Thermal Expansion 27 •• You need to fit a copper collar tightly around a steel shaft that has a diameter of 6.0000 cm at 20ºC. The inside diameter of the collar at that temperature is 5.9800 cm. What temperature must the copper collar have so that it will just slip on the shaft, assuming the shaft itself remains at 20ºC? Picture the Problem Because the temperature of the steel shaft does not change, we need consider just the expansion of the copper collar. We can express the required temperature in terms of the initial temperature and the change in temperature that will produce the necessary increase in the diameter D of the copper collar. This increase in the diameter is related to the diameter at 20°C and the increase in temperature through the definition of the coefficient of linear expansion. Express the temperature to which the copper collar must be raised in terms of its initial temperature and the increase in its temperature: TTT Δ+= i Apply the definition of the coefficient of linear expansion to express the change in temperature required for the collar to fit on the shaft: D DD D T αα Δ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ =Δ Substitute for ΔT to obtain: D D TT α Δ += i
• 5. Thermal Properties and Processes 413 Substitute numerical values and evaluate T: ( )( ) C220 cm5.9800K1017 cm9800.5cm0000.6 C20 16 °= × − +°= −− T The van der Waals Equation, Liquid-Vapor Isotherms, and Phase Diagrams 33 •• Using Figure 20-16, find the following quantities. (a) The temperature at which water boils on a mountain where the atmospheric pressure is 70.0 kPa, (b) the temperature at which water boils in a container where the pressure inside the container is 0.500 atm, and (c) the pressure at which water boils at 115ºC. Picture the Problem Consulting Figure 20-16, we see that: (a) At 70.0 kPa, water boils at approximately C90° . (b) At 0.500 atm (about 51 kPa), water boils at approximately C78° . (c) The pressure at which water boils at 115°C is approximately kPa127 . Conduction 35 • A 20-ft × 30-ft slab of insulation has an R factor of 11. At what rate is heat conducted through the slab if the temperature on one side is a constant 68ºF and the temperature of the other side is a constant 30ºF? Picture the Problem We can use its definition to express the thermal current in the slab in terms of the temperature differential across it and its thermal resistance and use the definition of the R factor to express I as a function of ΔT, the cross- sectional area of the slab, and Rf. Express the thermal current through the slab in terms of the temperature difference across it and its thermal resistance: R T I Δ = Substitute to express R in terms of the insulation’s R factor: ff / R TA AR T I Δ = Δ =
• 6. Chapter 20414 Substitute numerical values and evaluate I: ( )( )( ) h kBtu 2.1 Btu Ffth 11 F30F68ft30ft20 2 = °⋅⋅ °−° =I Radiation 41 • The universe is filled with radiation that is believed to be remaining from the Big Bang. If the entire universe is considered to be a blackbody with a temperature equal to 2.3 K, what is the λmax (the wavelength at which the power of the radiation is maximum) of this radiation? Picture the Problem We can use Wein’s law to find the peak wavelength of this radiation. Wein’s law relates the maximum wavelength of the background radiation to the temperature of the universe: T Kmm2.898 max ⋅ =λ Substituting for T gives: mm.31 K3.2 Kmm2.898 max = ⋅ =λ General Problems 49 •• The solar constant is the power received from the Sun per unit area perpendicular to the Sun’s rays at the mean distance of Earth from the Sun. Its value at the upper atmosphere of Earth is about 1.37 kW/m2 . Calculate the effective temperature of the Sun if it radiates like a blackbody. (The radius of the Sun is 6.96 × 108 m.). Picture the Problem We can apply the Stefan-Boltzmann law to express the effective temperature of the Sun in terms of the total power it radiates. We can, in turn, use the intensity of the Sun’s radiation in the upper atmosphere of Earth to approximate the total power it radiates. Apply the Stefan-Boltzmann law to relate the energy radiated by the Sun to its temperature: 4 r ATeP σ= ⇒ 4 r Ae P T σ = Express the surface area of the sun: 2 S4 RA π=
• 7. Thermal Properties and Processes 415 Relate the intensity of the Sun’s radiation in the upper atmosphere to the total power radiated by the sun: 2 4 R P I r π = ⇒ IRPr 2 4π= where R is the Earth-Sun distance. Substitute for Pr and A in the expression for T and simplify to obtain: 4 2 S 2 4 2 S 2 4 4 Re IR Re IR T σπσ π == Substitute numerical values and evaluate T: ( ) ( ) ( ) K5800 m106.96 Km W 105.671 m kW 1.35m101.5 4 28 2 8 2 211 = ×⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = − T 53 •• On the average, the temperature of Earth’s crust increases 1.0ºC for every increase in depth of 30 m. The average thermal conductivity of Earth’s crust material is 0.74 J/m⋅s⋅K. What is the heat loss of Earth per second due to conduction from the core? How does this heat loss compare with the average power received from the Sun (which is about 1.37 kW/m2 )? Picture the Problem We can apply the thermal-current equation to calculate the heat loss of Earth per second due to conduction from its core. We can also use the thermal-current equation to find the power per unit area radiated from Earth and compare this quantity to the solar constant. Express the heat loss of Earth per unit time as a function of the thermal conductivity of Earth and its temperature gradient: x T kA dt dQ I Δ Δ == (1) or x T kR dt dQ Δ Δ = 2 E4π Substitute numerical values and evaluate dQ/dt: ( ) kW103.1 m30 C1.0 Ksm J 0.74m1037.64 1026 ×=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅⋅ ×= π dt dQ Rewrite equation (1) to express the thermal current per unit area: x T k A I Δ Δ =
• 8. Chapter 20416 Substitute numerical values and evaluate I/A: ( ) 2 W/m0.0247 m30 C1.0 KsJ/m0.74 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ° ⋅⋅= A I Express the ratio of I/A to the solar constant: %002.0 kW/m1.37 W/m0.0247 constantsolar 2 2 < = AI 59 ••• A small pond has a layer of ice 1.00 cm thick floating on its surface. (a) If the air temperature is –10ºC on a day when there is a breeze, find the rate in centimeters per hour at which ice is added to the bottom of the layer. The density of ice is 0.917 g/cm3 . (b) How long do you and your friends have to wait for a 20.0-cm layer to be built up so you can play hockey? Picture the Problem (a) We can differentiate the expression for the heat that must be removed from water in order to form ice to relate dQ/dt to the rate of ice build-up dm/dt. We can apply the thermal-current equation to express the rate at which heat is removed from the water to the temperature gradient and solve this equation for dm/dt. In Part (b) we can separate the variables in the differential equation relating dm/dt and ΔT and integrate to find out how long it takes for a 20.0-cm layer of ice to be built up. (a) Relate the heat that must be removed from the water to freeze it to its mass and heat of fusion: fmLQ = ⇒ dt dm L dt dQ f= Using the definition of density, relate the mass of the ice added to the bottom of the layer to its density and volume: AxVm ρρ == Differentiate with respect to time to express the rate of build-up of the ice: dt dx A dt dm ρ= Substitute for dt dm to obtain: dt dx AL dt dQ ρf= The thermal-current equation is: x T kA dt dQ Δ =
• 9. Thermal Properties and Processes 417 Equate these expressions and solve for dt dx : x T kA dt dx AL Δ =ρf ⇒ x T L k dt dx Δ = ρf (1) Substitute numerical values and evaluate dt dx : ( )( ) ( )( ) cm/h0.70 cm/h0.697 h s3600 s m 94.1 m0.0100kg/m917 kg kJ 333.5 C10C0 Km W 0.592 3 = =×= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ °−−°⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = μ dt dx (b) Separating the variables in equation (1) gives: dt L Tk xdx ρf Δ = Integrate x from xi to xf and t′ from 0 to t: 'dt L Tk xdx tx x ∫∫ Δ = 0f f i ρ and ( ) t L Tk xx f 2 i 2 f2 1 ρ Δ =− ⇒ ( ) Tk xxL t Δ − = 2 2 i 2 ffρ Substitute numerical values and evaluate t: ( )( ) ( ) ( )[ ] d12 h24 d1 s3600 h1 s1003.1 m0.010m0.200 C10C0 Km W 0.5922 kg kJ 333.5 m kg 917 6 22 3 =×××= − °−−°⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =t