Chapter 01 Temperature and Heat (Pp 1-18)

CH 01

TEMPERATURE AND HEAT

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CHAPTER 01 TEMPERATURE AND HEAT 1-1 TEMPERATURE SCALES Problem 1-1 Absolute zero is − 273.15 0 C . Find absolute zero on the Fahrenheit scale. Solution The Fahrenheit and Celsius scales are related by TC TF − 32 = 5 9 0 Now TC = −273.15 C , therefore − 273.15 TF − 32 = 5 9 (−273.15)(9) = TF − 32 5 − 491.67 = TF − 32

TF = −491.67 + 32 = −459.67 0 F Hence absolute zero on the Fahrenheit scale is − 459.67 0 F . Problem 1-2 The melting point of tungsten (W) is 3410 0 C . Express this temperature on the Fahrenheit scale. Solution The Fahrenheit and Celsius scales are related by TF − 32 TC = 9 5 9 TF = TC + 32 5 9 TF = (3410) + 32 = 6170 0 F 5

CH 01 TEMPERATURE AND HEAT 2 Problem 1-3 Express the normal body temperature of 37 0 C on (a) the Fahrenheit scale and (b) the Kelvin scale. Solution (a) The Fahrenheit and Celsius scales are related by TF − 32 TC = 9 5 9 TF = TC + 32 5 9 TF = (37) + 32 = 98.6 0 F 5 (b) The Kelvin and Celsius scales are related by TK = TC + 273.15 TK = 37 + 273.15 = 310.15K Problem 1-4 The boiling point of nitrogen is 77.35 K . Express this temperature in degrees Fahrenheit. Solution As TC = TK − 273.15 TC = 77.35 − 273.15 = −195.80 C 9 Therefore TF = TC + 32 5 9 TF = (−195.8) + 32 = −320.44 0 F 5 Problem 1-5 At what temperature is the numerical value the same on the Kelvin and Fahrenheit scale? Solution The Fahrenheit and Kelvin scales are related by TF − 32 TK − 273.15 = 9 5

CH 01

TEMPERATURE AND HEAT 3 x − 32 x − 273.15 = 9 5 5 x − 160 = 9 x − 2458.35 2458.35 − 160 = 9 x − 5 x 4 x = 2298.35 2298.35 x= = 5750 F or 575 K 4 Problem 1-6 At what temperature is the Fahrenheit scale reading is equal to (a) twice that of the Celsius and (b) half that of the Celsius? Solution The Fahrenheit and Celsius scales are related by TC TF − 32 = 5 9 (a) Now TF = x and TC = 0.5 x therefore

0.5 x x − 32 = 5 9 (0.5 x)(9) = x − 32 5 0.9 x = x − 32 32 = x − 0.9 x 32 = 0.1 x 32 x= = 3200 F 0 .1 (b) Now TF = x and TC = 2 x therefore 2 x x − 32 = 5 9 18 x = 5 x − 160 18 x − 5 x = −160 13 x = −160

CH 01

TEMPERATURE AND HEAT 4 160 = −12.30 F x=− 13 Problem 1-7 At what temperature Celsius is the sum of Celsius temperature, the Kelvin temperature and the Fahrenheit temperature equal to 495.15 ? Solution Let x0C be the desired temperature, then x 0 C = ( x + 273.15) K TF − 32 TC and = 9 5 0 x F − 32 x = 9 5 9 x 0 F = x + 32 = 1.8 x + 32 5 0 x C + x in K + x 0 F = 495.15 Now x + ( x + 273.15) + (1.8 x + 32) = 495.15 x + x + 1.8 x = 495.15 − 273.13 − 32 3.8 x = 190 190 x= = 50 0 C 3 .8 Problem 1-8 The melting and boiling points of gold are 1064 0 C and 2660 0 C respectively. (a) Express theses temperature in Kelvins. (b) Compute the difference between these temperatures in Celsius degrees and Kelvins. Solution (a) TMELTING = 1064 0 C = (1064 + 273) K = 1337 K TBOILING = 2660 0 K = (2660 + 273) K = 2933 K

(b) TBOILING − TMELTING = 2660 − 1064 = 1596 0 C TBOILING − TMELTING = 2933 − 1337 = 1596 K

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1-2 LINEAR EXPANSION Problem 1-9 By how much would a 10 m long aluminum bar change its length in the process of going from 30 0 C to 50 0 C ? The temperature coefficient of linear expansion for aluminum is 2.5 × 10 −5 0 C −1 . Solution The change in length is given by ∆L = α L0 ∆T

∆L = (2.5 × 10 −5 )(10)(50 − 30) ∆L = 5 × 10 −3 m = 5 mm Problem 1-10 A 100 m steel measuring tape was marked and calibrated at 20 0 C . At a temperature of 30 0 C , what is percentage error in a 100 m distance when using this tape? The value of coefficient of linear expansion for steel is α = 1.2 × 10 −5 0 C −1 . Solution Now ∆L = α L0 ∆T ∆L = (1.2 × 10 −5 )(100)(30 − 20) ∆L = 0.012 m ∆L Percentage error = × 100 % L0 0.012 = ×100 = 0.012 % 100 Problem 1-11 An aluminum flagpole is 33 m high. By how much does its length increases as the temperature increases by 15 0 C ? The coefficient of linear expansion for aluminum is

α = 2.5 × 10 −5 0 C −1 .

CH 01 TEMPERATURE AND HEAT Solution The change in length is given by ∆L = α L0 ∆T

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∆L = (2.5 × 10 −5 )(33)(15) ∆L = 1.2375 × 10 −2 m ≅ 12 mm Problem 1-12 Steel railroad tracks are laid when the temperature is − 5 0 C . A standard section of rail is then 12.0 m long. What gap should be left between rail sections so that there is no compression when the temperature gets as high as 42.0 0 C ? The coefficient of linear expansion for steel is

α = 1.2 × 10 −5 0 C −1 . Solution The linear expansion of a section of steel for given temperature change is calculated as ∆L = α L0 ∆T ∆L = (1.2 × 10 −5 )(12.0){( 42.0) − (−5.0)} ∆L = 6.768 × 10 −3 m ≅ 6.8 mm Hence the desired gap between two rail sections is 6.8 mm . Problem 1-13 A circular hole in an aluminum plate is 2.725 cm in diameter at 20 0 C . What is its diameter when the temperature of plate is raised to 140 0 C ? Solution Now ∆L = α L0 ∆T

∆L = (2.5 × 10 −5 )(2.725)(140 − 20) ∆L = 0.008 cm The new diameter will be L1 = L0 + ∆L

L1 = 2.725 + 0.008 = 2.733 cm

CH 01 TEMPERATURE AND HEAT 7 Problem 1-14 A copper telephone wire has essentially no sag between poles 35.0 m apart on a winter day when the temperature is − 20.0 0 C . How much longer is the wire on a summer day when TC = 35.0 0 C ? The coefficient of linear expansion for copper is α = 1.7 × 10 −5 0 C −1 . Solution The increase in length is given by ∆L = α L0 ∆T ∆L = (1.7 × 10 −5 )(35.0){35.0 − (−20.0)} ∆L = 3.27 × 10 −2 m = 3.27 cm Problem 1-15 A brass rod 1.5 m long expands 1.89 mm when heated from 20 0 C to 90 0 C . Find the coefficient of linear expansion of brass. Solution The change in length is given by ∆L = α L0 ∆T

α=

1.89 × 10 −3 ∆L = 1.8 × 10 −5 0 C −1 = L0 ∆T (1.5)(90 − 20)

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1-3 SUPERFICIAL OR SURFACE EXPANSION Problem 1-16 A glass window is 200 m by 300 m at 10 0 C . By how much has its area increased when its temperature is 40 0 C ? Assume that the glass is free to expand. The coefficient of linear expansion for glass is 9 × 10 −6 0 C −1 . Solution The increase in surface area is given by ∆A = β A ∆T = 2 α A ∆T Θ β = 2 α ∆A = 2(9 × 10 −6 ){(200 × 10 −2 )(300 × 10 −2 )}(40 − 10) ∆A = 3.24 × 10 −3 m 2

Problem 1-17 A square hole 8.0 cm along each side is cut on a sheet of copper. (a) Calculate the change in the area of this hole if the temperature of the sheet is increased by 50.0 0 K . (b) Does this change represent an increase or a decrease in the area enclosed by the hole? Solution (a) The change in the area of the hole is given by ∆A = β A ∆T = 2 α A ∆T Θ β = 2 α ∆A = 2(1.7 × 10 −5 )(8.0 × 10 −2 ) 2 (50.0) ∆A = 1.088 × 10 −5 m 2 ∆A = 0.1088 × 10 −4 m 2 ≅ 0.109 cm 2 (b) As the length of each side of the square hole increases with increase in temperature, therefore the above ∆A represents an increase in the area of the hole.

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1-4 CUBICAL OR VOLUMETRIC EXPANSION Problem 1-18 A quartz sphere is 8.75 cm in diameter. What will be its change in volume if it is heated from 30 0 C to 200 0 C ? The coefficient of volume expansion for quartz is 1 × 10 −6 0 C −1 . Solution The change in volume is given by ∆V = γ V0 ∆T  4 4   8.75  ∆V = γ  π r 3  ∆T = (1 × 10 −6 ) (π )   3 3   2 

3

 (200 − 30) 

∆V = 5.96 × 10 −2 cm 3 = 5.96 ×10 −8 m 3 Θ 1 cm 3 = 1 × 10 −6 m 3 Problem 1-19 Find the change in volume of an aluminum sphere of 10 cm radius when it is heated from 0 0 C to 100 0 C . The coefficient

of linear expansion for aluminum is α = 2.5 × 10 −5 0 C −1 . Solution The change in volume is given by ∆V = γ V0 ∆T = 3 α V0 ∆T The volume of a sphere is given by 4 V0 = π r 3 3 Hence 4  ∆V = 3 α  π r 3  ∆T = 4 π α r 3 ∆T 3  −5 ∆V = 4(π )(2.5 × 10 )(10 × 10 −2 ) 3 (100 − 0) ∆V = 3.14 × 10 −5 m 3 = 31.4 cm3 Θ 1 m 3 = 1 × 10 6 cm 3

CH 01 TEMPERATURE AND HEAT 10 Problem 1-20 A bowl made of Pyrex glass is filled to the very brim with 100 cm 3 of water at 10 0 C . How much will overflow when the temperature of the filled bowl is raised to 50 0 C ? The coefficient of volume expansion for water is 2.07 × 10 −4 0 C −1 . Solution The change in volume is given by ∆V = γ V0 ∆T ∆V = (2.07 × 10 −4 )(100 × 10 −6 )(50 − 10) ∆V = 8.28 ×10 −7 m 3 = 0.828 cm3 Θ 1 m 3 = 1 × 10 6 cm 3 Problem 1-21 The mercury in a thermometer at 0 0 C has a volume of 0.50 cm 3 . What will be its volume at 100 0 C ? The coefficient of volume expansion of mercury is 1.82 × 10 −4 0 C −1 . Solution The new volume ‘V’ is given by V = V0 + ∆V = V0 + γ V0 ∆T V = V0 (1 + γ ∆T )

V = (0.50 × 10 −6 ){1 + (1.82 × 10 −4 )(100 − 0)} V = 5.091× 10 −7 m 3 = 0.5091 cm3 Θ 1 m 3 = 1 × 10 6 cm 3 Problem 1-22 What is the volume of a lead ball at − 12 0 C if its final volume at 160 0 C is 530 cm 3 ? The coefficient of linear expansion for lead is α = 2.9 × 10 −5 0 C −1 . Solution Let V1 and V2 be the volume of given lead ball at − 12 0 C and 160 0 C respectively, then V2 = V1 + ∆V1 = V1 + 3 α V1 ∆T = V1 (1 + 3 α ∆T )

CH 01

TEMPERATURE AND HEAT 11 V2 530 V1 = = (1 + 3 α ∆T ) 1 + 3(2.9 × 10 −5 ){160 − (−12)}

V2 = 522 cm3 Problem 1-23 The volume of a metal block increases by 0.15 % when it is heated through 50 0 C . Calculate the coefficient of linear expansion for this metal. Solution The change in volume is given by ∆V = γ V0 ∆T = 3 α V0 ∆T  ∆V   1 

 0.15  

1



  α =  =    V  0   3∆T   100   3 × 50 

α = 1×10 −5 0C −1 Problem 1-24 A brass cube has an edge length of 33.2 cm at 20 0 C . Find (a) the increase in surface area and (b) the increase in volume when it is heated to 75 0 C . The coefficient of linear expansion for brass is 1.9 × 10 −5 0 C −1 . Solution The initial surface area and volume of the glass cube are given by A0 = 6 x 2 = 6(33.2 × 10 −2 ) 2 = 6.613 × 10 −1 m 2 V0 = x 3 = (33.2 × 10 −2 ) 3 = 3.659 × 10 −2 m 3

(a)

∆A = β A0 ∆T = 2 α A0 ∆T

(b)

∆A = 2(1.9 × 10 −5 )(6.613 × 10 −1 )(75 − 20) ∆A = 1.382 × 10 −3 m 2 ∆V = γ V0 ∆T = 3 α V0 ∆T ∆V = 3(1.9 × 10 −5 )(3.659 × 10 −2 )(75 − 20) ∆V = 1.147 × 10 −4 m3

CH 01 TEMPERATURE AND HEAT 12 Problem 1-25 The coefficient of volume expansion of glycerin is 5.05 × 10 −4 0 C −1 . What will be the fractional change in its density for a 50 0 C rise in temperature? Solution The change in density is given by m m ∆ρ = ρ '− ρ = − V' V  1  − ∆V  1 ∆ρ = m  − =m   V0 + ∆V V0  V0 (V0 + ∆V )   m   ∆V  m ∆V ∆V  = − ρ 0 = −   2 V0 V0  V0   V0  γ V0 ∆T ∆ρ ∆V =− =− = −γ ∆T V0 V0 ρ0 ∆ρ = −(5.05 ×10 − 4 )(50) = −2.525 ×10 − 2

∆ρ ≅ −

ρ0

Problem 1-26 When the temperature of a metal cylinder is raised from 60 0 C to 100 0 C , its length increases by 0.092 % . (a) Find the percentage change in density. (b) Identify the metal. Solution (a) The fractional change in density is given by ∆ρ ∆V ∆L =− = −3 = −3(0.092) % = −0.276 % ρ V L (b) Now ∆L = α L ∆T  ∆L   1  (0.092 %) = 2.3 × 10 −5 0 C −1   =  L   ∆T  100 − 60 The given metal is aluminum.

α =

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1-5 SPECIFIC HEAT Problem 1-27 What is the specific heat of a metal if 135 kJ of heat is needed to raise 5.1 kg of the metal from 20 0 C to 30 0 C ? Solution Now Q = m C ∆T

Q 135 ×103 C= = = 2647 J kg −1 0C −1 m ∆T (5.1)(30 − 20) Problem 1-28 The temperature of a silver bar rises by 10 0 C when it absorbs 1.23 kJ of energy by heat. The mass of the bar is 525 g . Determine the specific heat of silver. Solution The heat absorbed is given by Q = m C ∆T Q 1.23 × 10 3 C= = = 234 J kg −1 0 C −1 −3 m ∆T (525 × 10 )(10) Problem 1-29 The brake linings of the wheels of a car have total mass 4.8 kg and specific heat capacity 1200 J kg −1 K −1 . Calculate the maximum possible temperature rise of the brake linings when the car (of mass 800 kg ) traveling at 15 m s −1 is brought to rest by applying the brakes. Solution Assuming no lasses we have Heat generated at brake linings = Kinetic energy of the car 1 m C ∆T = M v 2 2 M v2 (800)(15) 2 ∆T = = = 15.6 K 2 m C 2(4.8)(1200)

CH 01 TEMPERATURE AND HEAT 14 Problem 1-30 (a) Compute the possible increase in temperature for water going over Niagara fall, 49.4 m high. (b) What factors would tend to prevent this possible rise? (Given that C = 4190 J kg −1 K −1 for water) Solution (a) If the change in potential energy of water appears as rise in its internal energy, then m C ∆T = m g h

g h (9.8)(49.4) = = 0.116 K C 4190 (b) The water will evaporate during its fall accompanied by cooling, therefore the above rise in temperature i.e. 0.116 K is not observed. ∆T =

Problem 1-31 A 2 kW electric heater is used in a well-insulated hot water tank of capacity 200 litres to raise the temperature of water from 5 0 C to 60 0 C . How long will it take to change the temperature of water in the tank? Solution Now Q = P t = m C ∆T

m C ∆T ρ V C ∆T = P P 3 −3 (1 × 10 )(200 × 10 )(4190)(60 − 5) t= 2 × 10 3 t = 23045 s = 6 h 24 min 5 s t=

CH 01 TEMPERATURE AND HEAT 15 Problem 1-32 Calculate the minimum amount of heat required to completely melt 130 g of silver (melting point1235 K ) 0

initially at 16.0 C . The specific heat and latent heat of fusion of silver are 236 J kg −1 K −1 and 1.05 × 10 5 J K −1 respectively. Assume that specific heat does not change with temperature. Solution The desired amount of heat is given by Q = m C ∆T + m L f

Q = m (C ∆T + L f ) Now m = 130 g = 0.130 kg C = 236 J kg −1 K −1 L f = 1.05 × 10 5 J K −1 T1 = 16.0 0 C = (16 + 273) K = 289 K T2 = 1235 K ∆T = T2 − T1 = 1235 − 289 = 946 K Hence Q = (0.130){(236)(946) + (1.05 × 10 5 )} Q = 4.267 ×10 4 J = 42.67 kJ Problem 1-33 What mass of steam at 100 0 C must be mixed with 150 g of ice at 0 0 C , in a thermally insulated container, to produce liquid at 50 0 C ? The specific heat, latent heat of fusion and latent heat of vapourization of water are 4190 J kg −1 K −1 , 3.34 × 10 5 J K −1 and 2.256 × 10 6 J K −1 respectively.

CH 01 TEMPERATURE AND HEAT 16 Solution Let ‘m’ be the desired mass of steam, then heat librated by steam to produce water at 50 0C will be Q = m C ∆T + m Lv = m (C ∆T + Lv )

Q = m {(4190)(100 − 50) − (2.256 × 10 6 )} Q = m (2.4655 ×106 ) J The above heat will be absorbed by m1 = 150 g = 0.150 kg of ice to produce water at 50 0C. Hence Q = m1C ∆T + m1 L f = m1 (C ∆T + L f ) m(2.4655 × 10 6 ) = (0.150){( 4190)(50 − 0) + (3.34 × 10 5 )} (2.4655 × 10 6 )m = (8.1525 × 10 4 ) m=

8.1525 ×10 4 = 0.033 kg = 33 g 2.4655 ×106

Problem 1-34 How much energy is required to change a 40 g ice cube from ice at − 10 0 C to steam at 110 0 C ? Solution The net heat to complete the above change consists of the following processes. Q = Heat needed to convert − 10 0 C ice into 0 0 C ice + Heat needed to convert 0 0 C ice into 0 0 C water + Heat needed to convert 0 0 C water into 100 0 C water + Heat needed to convert 100 0 C water into 100 0 C steam + Heat needed to convert 100 0 C steam into 110 0 C steam.

Q = m C ICE ∆T1 + m L f + m CWATER ∆T2 + m Lv + m C STEAM ∆T3 Q = m(C ICE ∆T1 + L f + CWATER ∆T2 + m Lv + C STEAM ∆T3 ) Q = (0.040){(2090)(10) + (3.34 × 10 5 ) + (4190)(100) + (2.256 × 10 6 ) + (92010)(10)} Q = 1.22 ×105 J

CH 01

TEMPERATURE AND HEAT 17 1-6 THERMAL CONDUCTION Problem 1-35 Calculate the rate at which heat would be lost on a very cold winter day through a 6.2 m × 3.8 m brick wall 32 cm thick. The inside temperature is 26 0 C and the outside temperature is − 18 0 C ; assume that the thermal conductivity of the brick is 0.74 W m −1 K −1 . Solution The quantity of heat ‘Q’ transferred from one face to the other is given by ∆T Q = −k A ∆t ∆x Q ∆T H= = −k A ∆t ∆x −1 −1 Now k = 0.74 W m K A = 6.2 × 3.8 m 2 = 23.56 m 2 ∆T = T1 − T2 = −18 − 26 = −44 0 C (Note that the difference in temperature will be same on Celsius and Kelvin scales) ∆x = 32 cm = 0.32 m Hence  − 44  3 H = −(0.74)(23.56)  = 2.397 × 10 W  0.32  Problem 1-36 A steel plate measures 20 cm by 20 cm and is 5.0 mm thick. One face is maintained at140 0 C . How much power would be needed to maintain the other face at 159 0 C ? The value of thermal conductivity for steel is 46 W m −1 K −1 . Solution The desired power is given by ∆T H = −k A ∆x −1 −1 Now k = 46 W m K

CH 01 TEMPERATURE AND HEAT 18 2 2 −4 2 A = 20 × 20 cm = 400 cm = 400 × 10 m = 4 × 10 −2 m 2 ∆T = T1 − T2 = 140 − 150 = 10 0 C = 10 K ∆x = 5.0 mm = 5 × 10 −3 m Hence  − 10  H = −(46)(4 × 10 −2 ) = 3680 W −3   5 × 10  Problem 1-37 How thick a concrete wall would be needed to give the same insulating value as 10 cm of fiberglass? The values of thermal conductivity of concrete and fiberglass are 1 W m −1 K −1 and 0.042 W m −1 K −1 respectively. Solution Now H FIBERGLASS = H CONCRETE ∆T ∆T − kF A = −k C A ∆x F ∆xC k  ∆xC =  C  ∆x F  kF   1  ∆xC =   (10) = 238 cm = 2.38 m  0.042 