Chapter 05 Entropy (Pp 100-118)

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CHAPTER 05 ENTROPY 5-1 INTRODUCTION Example 5-1 Calculate the change in entropy when 25 kJ of energy is transferred reversibly and isothermally as heat to a large block of ice at (a) 0 0 C and (b) 100 0 C . Solution (a) The change in entropy is defined as ∆Q ∆S = T 25 × 10 3 = 92 J / K ∆S = 0 + 273 (b) The change in entropy is defined as ∆Q ∆S = T 25 × 10 3 ∆S = = 67 J / K 100 + 273 Example 5-2 An ideal gas undergoes a reversible isothermal expansion at 132 0 C . The entropy of the gas increases by 46.2 J / K . How much heat is absorbed? P.U. B.Sc. 2007 Solution The change in entropy is defined as ∆Q ∆S = T ∆Q = (∆S ) T = (46.2)(132 + 273) ∆Q = 1.871×10 4 J / K = 18.71 kJ / K

CH 05 ENTROPY 101 Example 5-3 The ends of a well insulated metal rod are made in contact with two reservoirs at 60 0 C and − 15 0 C . During certain time 200 J of heat are transferred from hot reservoir to cold reservoir. What are the entropy changes at the two reservoirs and net change in entropy? K.U. B.Sc. 2000 Solution The entropy change in the hot reservoir is given by ∆QH 200 ∆S H = =− = −0.601 J / kg TH (60 + 273) The entropy change in the cold reservoir is given by ∆QL 200 ∆S L = = = 0.775 J / kg TL (−15 + 273) The net change in the entropy of the system is ∆S = ∆S H + ∆S L ∆S = −0.601 + 0.775 = 0.174 J / kg Example 5-4 A 1500 kg car traveling at 100 km / h crashes into a concrete wall. If the temperature of the air is 20 0 C , calculate the entropy change in the universe. Solution The entropy change is defined as ∆Q (m v 2 / 2) ∆S = = T T Now m = 1500 kg (100)(1000) 250 v = 100 km / h = m/s = m/s 3600 9 T = 200 C = (20 + 273) K = 293 K Therefore (1500)(250 / 9) 2 ∆S = = 1975 J / K 2(293)

CH 05 ENTROPY 102 Example 5-5 250 g of ice melts (reversible) to water, the temperature remaining at 0 0 C . Calculate the entropy change for ice if its heat of fusion is 334 kJ / kg . P.U. B.Sc. 2004, 2008 Solution For an isothermal and reversible process, the change in entropy is given by ∆Q m L f ∆S = = T T (250 ×10 −3 )(334 ×103 ) ∆S = = 306 J / K 273 Example 5-6 A lump of ice whose mass is 235 g melts (reversibly) to water, the temperature remaining at 0 0 C throughout the process. What is the entropy change for the ice? The heat of fusion of ice is 334 kJ / kg . K.U. B.Sc. 2004 Solution The change in entropy is given by ∆Q m L f ∆S = = T T −3 Now m = 235 g = 235 × 10 kg = 2.35 × 10 −1 kg L f = 334 kJ / K = 334 × 10 3 J / K = 3.34 × 10 5 J / kg T = 0 0 C = 273 K

Hence ∆S =

(2.35 × 10 −1 )(3.34 × 10 5 ) = 288 J / K 273

Example 5-7 Calculate the change of entropy when 54.6 g of ice at 0 0 C is converted into water at 0 0 C at normal pressure. The latent heat of fusion of ice is 80 calories per gram. B.U. B.Sc. 1997A

CH 05 ENTROPY 103 Solution The change in entropy at constant temperature is given by ∆Q m L f ∆S = = T T (54.6)(80) ∆S = calories / K 273 (54.6)(80)(4.186) ∆S = J / K = 66.976 J / K 273 Θ 1 calorie = 4.186 J Example 5-8 Calculate the change of entropy when 27.3 g of ice at 0 0 C is converted into water at 0 0 C at normal pressure. The latent heat of fusion of ice is 80 calories per gram. K.U. B.Sc. 1999 Solution The change in entropy at constant temperature is given by ∆Q m L f ∆S = = T T Now m = 27.3 g = 27.3 × 10 −3 kg = 2.73 × 10 −2 kg

L f = 80 calories / gram (80)(4.186) J / kg = 3.3488 × 10 5 J / kg 10 −3 T = 0 0 C = 273 K

Lf =

Hence (2.73 × 10 −2 )(3.3488 × 10 5 ) ∆S = = 33.488 J / kg 273 Example 5-9 An ice tray contains 500 g of liquid water at 0 0 C . Calculate the change in entropy of water as it freezes slowly and completely at 0 0 C .

CH 05 ENTROPY 104 Solution For a freezing process m Lf ∆Q ∆S = =− T T (500 ×10 −3 )(3.34 ×105 ) ∆S = − = −612 J / K 273 Example 5-10 Calculate the change in entropy when 10 kg of water at 100 0 C is converted into steam at same temperature. The latent heat of vapourization for water is 2.256 × 10 6 J / kg . Solution Now ∆Q m Lv ∆S = = T T (10)(2.256 ×106 ) ∆S = = 6.05 ×10 4 J / K 100 + 273 Example 5-11 Calculate the change of entropy when 7.46 grams of water at 100 0 C are converted into steam at 100 0 C at normal pressure. The latent heat of vapourization for water at 100 0 C is 537 calories/gram. B.U. B.Sc. (Hons.) 1984A Solution The entropy change is defined as ∆Q m Lv ∆S = = T T (7.46)(537) ∆S = = 10.74 calories / K 100 + 273 ∆S = (10.74)(4.186) J / K = 44.96 J / K

Θ 1 calorie = 4.186 J

CH 05 ENTROPY 105 Example 5-12 The heat of vapourization of a compound at 27 0 C is found to be 29.288 kJ mol −1 . Calculate the mole entropy of vapourization at 27 0 C . K.U. B.Sc. 1999 Solution The entropy change is defined as ∆Q m Lv ∆S = = T T (1)(29.288 × 10 3 ) ∆S = = 97.6 J / K (27 + 273) Example 5-13 The melting point of copper is 2840 K . What will be the change in entropy when 20 kg of copper melts? The heat of vapourization of copper is 4.73 × 10 6 J kg −1 . Solution The entropy change is defined as ∆Q m Lv ∆S = = T T (20)(4.73 × 10 6 ) ∆S = = 3.331 × 10 4 J / K 2840 Example 5-14 Calculate the change in entropy when 25 g of steam at 100 0 C is converted into water at the same temperature. The latent heat of vapourization for water is 2.256 × 10 6 J / kg . Solution m Lv ∆Q ∆S = =− T T −3 (25 ×10 )(2.256 ×106 ) = −151 J K −1 ∆S = − (100 + 273) The negative sign indicates that the entropy of the steam decreases.

CH 05 ENTROPY 106 Example 5-15 What is the entropy change of 0.500 kg of steam at 1.00 atm pressure and 100 0 C when it condenses to 0.500 kg of water at 100 0 C ? Solution As steam condenses to water, therefore entropy change will be negative and given by m Lv ∆Q ∆S = − =− T T (0.500)(2.256 × 10 6 ) ∆S = − = −3.024 × 10 3 J K −1 (100 + 273)

Example 5-16 For the Carnot cycle shown in Figure, calculate (a) the heat that enters and (b) the work done on the system.

Solution (a) Heat enters the system along the top path and given by Qin = TH ∆S Qin = (400)(0.60 − 0.10) = 200 J (b) Heat leaves the system along the bottom path and given by

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ENTROPY Qout = TL ∆S

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Qout = (250)(0.10 − 0.60) = −125 J For a cyclic path Q +W = 0 W = −Q = −{Qin − Qout } = −(200 − 125) = −75 J

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5-2 ENTROPY CHANGE FOR AN IDEAL GAS 5-2(A) ISOTHERMAL PROCESS Example 5-17 An ideal gas undergoes an isothermal expansion at 77 0 C increasing its volume from 1.3 to 3.4 L . The entropy change of the gas is 24 J / K . How many moles of gas are present? Solution The heat added to the system, ∆Q , will be equal to the work done by the gas in the form of isothermal expansion and is given by Vf  ∆Q = n R T λn    Vi 

Vf  ∆Q = n R λn   T  Vi  24 = n (8.314) λn (3.4 / 1.3) 24 n= = 3.003 mol (8.314) λn (3.4 / 1.3) Example 5-18 One mole of an ideal gas expands isothermally to ten times its initial volume. Calculate the entropy change. Solution The entropy change is defined as Vf  ∆Q ∆S = = n R λn   T  Vi  ∆S =

∆S = (1)(8.314) λn (10) = 19.1 J / K Example 5-19 Two moles of an ideal gas undergoes a reversible isothermal expansion from 0.028 m 3 to 0.042 m 3 at a temperature of 25 0 C . What is the change in entropy of the gas?

CH 05 ENTROPY Solution The entropy change is defined as ∆Q ∆S = = n R λn T

109

Vf     Vi   0.042  ∆S = (2)(8.314) λn   = 6.742 J / K  0.028  Example 5-20 Five moles of an ideal gas are allowed to expand isothermally at 27 0 C from a pressure of 20 atm to 4 atm . Calculate the entropy change. Solution The entropy change at constant temperature and in terms of initial and final pressures is given by  p  ∆S = n R λn  i  p   f  ∆S = (5)(8.314) λn (20 / 4) = 60.9 J / K Example 5-21 A sample of nitrogen gas of mass 25 g at 250 K and 2 atm expands isothermally until its pressure is 4.3 atm . Calculate the change in entropy of the gas. The molar mass of nitrogen is 28 g / mol . Solution The desired change in entropy is given by  p  ∆S = n R λn  i  p   f 

 25   21.1  ∆S =   (8.314) λn   = 16.5 J / K  28   4 .3 

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5-2 ENTROPY CHANGE FOR AN IDEAL GAS 5-2(B) ISOBARIC PROCESS Example 5-22 One kilogram of water 0 0 C is heated to 100 0 C . Find the change in entropy. Specific heat of water is 4200 J kg −1 K −1 . B.U. B.Sc. 1986A Solution The change in entropy is given by  Tf  ∆S = m C λn    Ti  Now m = 1 kg

C = 4200 J kg −1 K −1 Ti = 0 0 C = 273 K T f = 1000 C = (100 + 273) K = 373 K

Hence ∆S = (1)(4200) λn (373 / 273) = 1311 J / kg Example 5-23 Calculate the entropy increase when 5 kg of water from 20 0 C to 80 0 C . Given that specific heat may be assumed to have the constant value of 4.91 × 10 3 J kg −1 K −1 . Solution The change in entropy is given by  Tf  ∆S = m C λn    Ti   80 + 273  3 ∆S = (5)(4.19 × 10 3 ) λn   = 3.902 × 10 J / kg  20 + 273 

CH 05 ENTROPY 111 Example 5-24 Calculate the change in entropy when a body of mass 5 mg is heated from 100 0 C to 300 0 C . Given specific heat of the body = 0.1 cal g −1 K −1 . K.U. B.Sc. 2006 Solution The change in entropy is given by  Tf  ∆S = m C λn    Ti 

Now

m = 5 mg = 5 × 10 −3 g = 5 × 10 −6 kg (0.1)(4.186) C = 0.1 cal g −1 K −1 = J kg −1 K −1 10 −3 C = 4.186 × 10 2 J kg −1 K −1 Ti = 100 0 C = (100 + 273) K = 373 K T f = 3000 C = (300 + 273) K = 573 K ∆S = (5 × 10 −3 )(4.186 × 10 2 ) λn (573 / 373) ∆S = 0.899 J / kg

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5-2 ENTROPY CHANGE FOR AN IDEAL GAS 5-2(C) ISOCHORIC PROCESS Example 5-25 Two moles of a monoatomic ideal gas is warmed slowly from 300 K to 400 K at constant volume. Determine the entropy change of the gas. Solution The entropy change at constant volume is given by  Tf  ∆S = n C v λn    Ti 

 Tf  3 3 n R λn   Θ For monoatomic gas C v = R 2 2  Ti  3 ∆S = (2)(8.314) λn (400 / 300) = 7.175 J / K 2 Example 5-26 An aluminum block of 10 kg mass is heated from 300 K to 800 K . Calculate the change in entropy if −1 −1 C v = 900 J kg K for aluminum. ∆S =

Solution The desired change in entropy is given by  Tf  ∆S = m C v λn    Ti 

∆S = (10)(900) λn (800 / 300) = 8.827 × 103 J / K

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5-3 MISCELLANEOUS Example 5-27 100 gms of ice at 0 0 C is heated to 100 0 C . Calculate the change in entropy if the specific heat of water is 4200 J /(kg • 0 C ) . B.U. B.Sc. (Hons.) 1987A Solution The increase in entropy during melting of 100 g of ice at 0 0 C

into water at 0 0 C is given by ∆Q1 m L f ∆S1 = = T1 T1 (0.100)(3.34 × 105 ) = 122.3 J / K 273 When 100 g of water is heated from 0 0 C to 100 0 C , the increase in entropy is given by T  ∆S 2 = m C λn  2   T1  ∆S1 =

 100 + 273  ∆S 2 = (0.100)(4200) λn   = 131.1 J / K  273  The net change in entropy is given by ∆S = ∆S1 + ∆S 2 ∆S = 122.3 + 131.1 = 253.4 J / K

Example 5-28 Calculate the change of entropy when 10 kg of ice at 0 0 C is changed into water at 2 0 C . (Heat of fusion of ice = 3.34 × 10 5 J / K and Specific heat of water = 4200 J /(kg • K ) ) K.U. B.Sc. 2001, 2003

CH 05 ENTROPY 114 Solution When 10 kg of ice at 0 0 C is melted into water at 0 0 C , the increase in entropy is given by ∆Q1 m L f ∆S1 = = T1 T1

(10)(3.34 × 10 5 ) = 1.223 × 10 4 J / K 273 When 10 kg of water is heated from 0 0 C to 2 0 C , the increase in entropy is given by T  ∆S 2 = m C λn  2   T1  ∆S1 =

 2 + 273  2 ∆S 2 = (10)(4200) λn   = 3.066 × 10 J / K  273  The desired entropy change is ∆S = ∆S1 + ∆S 2 ∆S = 1.223 × 10 4 + 3.066 × 10 2 = 1.254 × 10 4 J / K Example 5-29 200 g of aluminum at 107 0 C is mixed with 50 g of water at

17 0 C . Calculate the equilibrium temperature and change of entropy of the system. (Specific heats of aluminum and water are 910 J kg −1 K −1 and 4200 J kg −1 K −1 ) K.U. B.Sc. 2002 Solution Now m1 = 200 g = 0.2 kg

T1 = 107 0 C = (107 + 273) K = 380 K C1 = 910 J kg −1 K −1 m2 = 50 g = 0.05 kg T2 = 17 0 C = (17 + 273) K = 290 K C 2 = 4200 J kg −1 K −1 Let T be the final temperature of the mixture, then

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ENTROPY 115 Heat lost by aluminum = Heat gained by water m1C1 (T1 − T ) = m2 C 2 (T − T2 ) (0.2)(910)(380 − T ) = (0.05)(4200)(T − 290)

69160 − 182 T = 210 T − 60900 69160 + 60900 = 210 T + 182 T 130060 = 392 T 130060 T= = 332 K = 590 C 392 The change of entropy when the temperature 50 g of water rises from T2 to T is given by

T  ∆S1 = m2 C 2 λn    T2  ∆S1 = (0.05)(4200) λn (332 / 290) = 28.4 J / K The positive sign indicates gain in entropy. Similarly the change of entropy in 200 g of aluminum, when its temperature falls from T1 to T , is given by T  ∆S 2 = m1C1 λn    T1  ∆S 2 = (0.2)(910) λn (332 / 380) = −24.6 J / K The negative sign indicates the loss in entropy. Hence the change of entropy of the system is ∆S = ∆S1 + ∆S 2 ∆S = 28.4 − 24.6 = 3.8 J / K Example 5-30 At very low temperatures, the molar specific heat of many solids is (approximately) proportional to T 3 , that is, C v = A T 3 where ' A' depends on particular substance. For aluminum we have A = 3.15 × 10 −4 J /(mol • K 4 ) . Find the entropy change of 4.8 mol of aluminum when its temperature is raised from 5.0 K to 10 K .

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CH 05 ENTROPY Solution The entropy change is defined as T2 T dQ 2 m C v dT ∆S = ∫ =∫ T T T1 T1 T2

∆S =

m A T 3 dT =m A ∫T T 1

T2

∫T

2

dT

T1

T2 1 1 ∆S = m A T 3 T1 = m A T23 − T13 3 3 1 ∆S = (4.8)(3.15 ×10 − 4 ) (10) 3 − (5)3 = 0.441 J / K 3

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CH 05 ENTROPY 117 ADDITIONAL PROBLEMS (1) Calculate the change of entropy when 37.3 gms of water at 100 0 C are converted into steam at same temperature at normal pressure. The latent heat of steam of steam at 100 0 C is 537 calories per gm. B.U. B.Sc. 2009A 0 (2) 100 g of ice at 0 C is heated to 100 0 C . Find the change in entropy if the specific heat of water is 4200 J /(kg • K ) . B.U. B.Sc. (Hons.) 1987A (3) Calculate the change in entropy when 37.3 grams of water at 100 0 C are converted into steam at 100 0 C . The latent heat of steam at 100 0 C is 537 calories per gram. B.U. B.Sc. (Hons.) 1988A (4) Twenty-five kilograms of water at 15 0 C is heated to 100 0 C . Calculate the change in entropy if the specific heat of water is 4200 J /(kg • K ) . B.U. B.Sc. 1999A (5) Calculate the change of entropy when 27.3 grams of ice at 0 0 C are converted into water at 0 0 C , at normal pressure. The latent heat of fusion of ice is 80 calories per gram. K.U. B.Sc. 1999 0 (6) 10 kg of ice at 0 C is heated to 100 0 C . Find the change in entropy if the specific heat of water is B.U. B.Sc. 2001A 4200 J /(kg • K ) . (7) Calculate the change in entropy when 50 g of ice at 0 0 C changes to water at 2 0 C . (Heat of fusion of ice = 3.34 × 10 5 J / kg and specific heat of water = 4200 J /(kg • K ) . K.U. B.Sc. 2005 (8) An ideal gas is confined to a cylinder by a piston. The piston is pushed down slowly so that the gas temperature remains at 20 0 C . During the compression 730 joules of work is done on the gas. Find the entropy change of the gas. K.U. B.Sc. 2007

CH 05 ENTROPY 118 (9) Calculate the change of entropy when 27.3 gms of water at 0 0 C are converted into steam at same temperature at normal pressure. The latent heat of steam of steam at 100 0 C is 537 calories per gm. B.U. B.Sc. 2009S (10) Calculate the change of entropy when 250 gm of ice is melted at 0 0 C . Given that 3.34 × 10 5 joules of energy is required to melt 1 kg of ice at 0 0 C . B.U. B.Sc.(Hons.) 1988A Answers (1) 3.7 calories / K or 15.49 J / K (3) 53.7 calories / K or 224.79 J / K (5) 8 calories / K or 33.5 J / K (7) 631.52 J / K (8) 2.49 J / K (10) 306 J

(2) 253.4 J / K (4) 2.715 × 10 4 J K −1 (6) 1.311 × 10 4 J / K (9) 8 calories/K