G 10 thermal energy, Temperature and heat original FINAL.pptx

physics

1. Thermal Energy Temperature & Heat Grade 10 Physics - Understanding the Relationship Between Temperature and Kinetic Energy 1

2. Remember The atoms and molecules that make up matter are in continuous, random motion. Matter in Motion Matter is made of tiny particles—atoms and molecules. Particles are in constant, random motion Faster = More KINETIC energy Particles in hot objects move faster than cooler objects. 2

3. Thermal Energy Kinetic energy + potential energy of an object = thermal energy of that object. 3

4. 1. Thermal energy relationships a. Depends on temperature, mass, and type of substance b. As temperature increases, so does thermal energy (because the kinetic energy of the particles increased). c. Even if the temperature doesn’t change, the thermal energy of a more massive substance is higher (because it is a total measure of energy). 4

5. Which beaker of water has more thermal energy? B - same temperature, more mass 200 mL 80ºC A 400 mL 80ºC B 5

6. Temperature and Heat Are temperature and heat the same? 6

7. Temperature and Heat Are temperature and heat the same? No, because a spoonful of boiling water (100 o C) will have less thermal (heat) energy …. 7

8. Temperature and Heat Are temperature and heat the same? No, because a spoonful of boiling water (100 o C) will have less thermal (heat) energy …. … than a beaker of boiling water (at the same temperature). 8

9. Temperature is a measure of the average kinetic energy of the individual particles in a substance. 9

10. A. Temperature Temperature is measured with a thermometer and can be measured in Kelvin, Celsius, and Fahrenheit Absolute zero-temperature at which particles stop moving 0 o K 10

11. SI unit for temp. is the Kelvin a. K = C + 273 (50 C = ______K) b. C = K – 273 (50 K = ______C) Example 1 Convert 500 K to celsius : Given, kelvin temperature = 500 K Example 2 Convert 750 K to celsius : Given, kelvin temperature = 750 K 11

12. Question: Ethyl alcohol boils at 78.5 o C and freezes at -117 o C under a pressure of 1 atm. Convert these temperatures to Kelvin scale. 12

13. What is absolute zero on the kelvin scale? Although obtaining negative values for temperature on the Celsius scale is perfectly natural, the Kelvin scale has a minimum value of zero. Zero Kelvin is also called absolute zero. It is the point at which no more  molecular motion , and there is no chance of lower temperature. Therefore, this implies that the lowest Celsius temperature that can possibly be achieved is equal to minus 273° C. 13

14. Temperature and Particle Motion 14

15. Interactive Activity Observe the motion of particles in the simulation at different temperatures. Discuss: - How does particle motion change with temperature? Phet simulation 1 Phet simulation 2 - What happens as temperature approaches absolute zero? 15

16. Quick Assessment: MCQs 1. What does temperature measure? A. Total energy of particles B. Average kinetic energy of particles C. Potential energy of particles 2. What happens to particle motion as temperature increases? A. Particles move slower B. Particles move faster C. Particles stop moving 16

17. Quick Assessment: MCQs 1. What does temperature measure? A. Total energy of particles B. Average kinetic energy of particles C. Potential energy of particles Answer: B 2. What happens to particle motion as temperature increases? A. Particles move slower B. Particles move faster C. Particles stop moving Answer: B 17

18. Temperature Thermal Energy Heat Th e t r a n s f e r o f e n e r gy b e t w e e n o b j e c t s t h a t a r e a t d i f f e r e n t t em p e r a t u r e s The total i nterna l energ y of molecules A measure of the average kinetic energy of the particles substance in a Degrees Fahrenheit , d e g r e e s C e l s i u s , Kelvins Joules Jou l es or Var i es w i th mass, s p e c i f i c h e a t c a pa c i ty, an d t e m p . c h a n g e Var i es with mass and temp . Does NOT vary with mass T hermal is ach i eved when all substance have reached t he same E quilibrium temperature. There is no mo r e therma l ene r gy transfer. 18

19. Recap and Homework Recap: - Thermal energy is the total kinetic energy of particles. - Temperature measures average kinetic energy. - Absolute zero is the minimum kinetic energy state. Homework: Research a real-world application of temperature and kinetic energy, and prepare a short presentation. 19

20. Heat a. The Flow of thermal energy from one object to another. b. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler Cup gets cooler while hand gets warmer 20

21. Heat The movement of heat from a high temperature to low temperature – when two substances at different temperatures are mixed together, heat flows from the warmer body to the cooler body until they reach the same temperature or equilibrium (Zeroth Law of Thermodynamics – Thermal Equilibrium). 21

22. Heat transfer only when there is a difference in temperature between to objects 22

23. How does heat transfer ? Heat transfer by three ways: Conduction Convection Radiation 23

24. A. How is heat transferred? What type of HEAT TRANSFER is occurring in the pictures? Conduction, convection or radiation? CONDUCTION – The transfer of thermal energy with NO transfer of matter. Occurs because particles that make up matter are in constant motion and have collisions 24

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27. Heat transfer by convetion Learning objective: Describe the three modes of heat transfer: convection and radiation, Provide examples of each By the end of the lesson: Students can describe the processes of convection and radiation and provide basic examples of each. By explain in writing how a campfire transfers heat through radiation and how a pot of boiling water demonstrates convection. 27

28. HEAT TRANSFER What type of HEAT TRANSFER is occurring in the pictures? Conduction, convection or radiation? CONVECTION – The transfer of thermal energy when particles of a liquid or gas move from one place to another 28

29. How convection occur ? • As the water above the flame heats up, • it expands, • becomes less dense, • and is pushed upwards, • while the cooler, more dense water sinks to take its place. 29

30. Convection Currents Stream of warm moving fluids are called convection currents 30

31. Applications on convection (The heater) What about the Air conditioner! 31

32. Convection currents inside Earth Tectonic drift 32

33. HEAT TRANSFER Convection currents – in the earth and sun The circular flow of hot and cold creates convection currents 33

34. HEAT TRANSFER What type of HEAT TRANSFER is occurring in the pictures? Conduction, convection or radiation? RADIATION – The transfer of thermal energy by electromagnetic waves (IR) moving through ( vacuum ) space. ALL OBJECTS radiate energy! 34

35. Radiation Heat radiation is the infra-red radiation 35

36. Check point Heat transfer by radiation is not possible from human beings to their environment. does not occur from light bulbs - they are too bright. does not require any material between the radiator and the object receiving the radiation. none of the above. 36

37. Check point Heat transfer by radiation a) is not possible from human beings to their environment. b) does not occur from light bulbs - they are too bright. *c) does not require any material between the radiator and the object receiving the radiation. d) none of the above. 37

38. Absorption and reflection Absorbtion experiment When an object is heated by an external source of heat by radiation, The amount of heat reflected and absorbed heat radiation ( infrared ) varies according to the color of the surface of the objects. 38

39. Check point Which object cools fastest at night? a) White rock b) Black soil c) Silver car d) Green grass 39

40. Check point Which object cools fastest at night? a) White rock *b) Black soil c) Silver car d) Green grass 40

41. Emission Emission experiment When an object is hotter than surround it radiates infrared ( emits heat ) to the surround. the amount of radiated ( emitted ) heat radiation ( infra-red ) also varies according to the color of the surface . 41

42. Two objects at the same temperature but different colors will: a) Emit the same radiation b) Emit different radiation c) Absorb the same heat d) Reflect the same light 42

43. Check point Two objects at the same temperature but different colors will: a) Emit the same radiation *b) Emit different radiation c) Absorb the same heat d) Reflect the same light 43

44. Factors affect emission • If the surface area of the container increased , the amount of emitted heat radiation will also increase . • If objects have the same temperature and same color but different surface areas , the greater surface area will emit larger amount of heat radiation and cools faster . Black color is good emitter (radiator) of heat radiation (infrared). White color is bad emitter (radiator) heat radiation. 44

45. Summary Dark and dull surfaces are ________ absorbers and ________ emitters of infra-red radiation. Light and shiny surfaces are ________ absorbers and ________ emitters of infra-red radiation, they are ________ reflectors . 45

46. Summary Dark and dull surfaces are GOOD absorbers and GOOD emitters of infra-red radiation. Light and shiny surfaces are POOR absorbers and POOR emitters of infra-red radiation, they are GOOD reflectors . 46

47. Application • A vacuum or Thermos flask keeps hot liquids hot or cold liquids cold • It is very difficult for heat to travel into or out of the flask 47

48. Preventing losing heat by conduction The flask is made of a double-walled vessel with vacuum between the walls Vacuum is a good insulator so there is no heat conduction between the hot water inside and the cold air outside. Preventing losing heat by radiation Radiation is reduced by plating the walls with silvery color from inside so heat radiation that is emitted from the hot liquid, will reflect back again Preventing losing heat by convection The hot water vapour that rises up by convection is prevented from going out by the stopper, Therefore, the liquid will be hot as long as possible. 48

49. Exit ticket 49

50. Bell Ringer 50

51. Heat Transfer Video https://www.generationgenius.com/videolessons/heat-transfer-of-thermal-energy-video-for-kids/?gclid=CjwKCAjwhaaKBhBcEiwA8acsHCME7fLqjSl5OjwqwJnWjmKglMa5ezXQ4I-fTz-ae_hClgEUU2Vh5hoCk34QAvD_BwE 51

52. Heat Transfer 1.Specific Heat Capacity (C p ) amount of energy required to raise the temp. of 1 kg of material by 1 Kelvin units: J/( kg·K ) or J/(kg·°C) or J/(g·°C) MUST PAY ATTENTION TO UNITS 52

53. Heat Transfer Which sample will take longer to heat to 100°C? 50 g Al 50 g Cu 53

54. Heat Transfer Which sample will take longer to heat to 100°C? 50 g Al 50 g Cu Al - It has a higher specific heat. Al will also take longer to cool down. 54

55. What is specific heat capacity? This is a measure of how much energy a material can store. It is the energy needed to raise the temperature of 1kg of a material by 1 o C. It is measured in J/kg o C If we heated these materials for 10 minutes , which would get hottest? copper limestone water http://www.youtube.com/watch?v=D3CwpfBzF94 55

56. SHC of different materials The copper became hottest because it has the lowest SHC. The water has the highest SHC as it absorbed the energy without becoming very hot. 82 o C 65 o C 43 o C 56

57. Heat Transfer Q = m   T  C p Q : heat (J) m : mass (kg)  T : change in temperature (°C) C p : specific heat (J/kg·°C)  T = T f - T i – Q = heat loss + Q = heat gain 57

58. m = Δ Q c x T T = Δ Q c x m c = Δ Q c x T Rearranging the equation: Energy transferred (J) mass kg temperature change o C specific heat capacity J/kg o C x x 58

59. m = Δ Q c x Δ T Δ T = Δ Q c x m c = Δ Q c x Δ T Rearranging the equation: Energy transferred (J) mass kg temperature change o C specific heat capacity J/kg o C x x 59

60. m = Δ E c x Δ θ Δ θ = Δ E c x m c = Δ E c x Δ T Rearranging the equation: Energy transferred (J) mass kg temperature change o C specific heat capacity J/kg o C x x 60

61. Practice Problem What is the specific heat of silver if the temperature of a 15.4 g sample of silver is increased from 20.0 o C to 31.2 o C when 40.5 J of heat is added? 61

62. Givens: m = 15.4 g T i = 20.0 o C T f = 31.2 o C Q = 40.5 J Q = mC ∆T 40.5=15.4(C)(31.2-20.0) 40.5=15.4(C)(11.2) 40.5=172.48(C) C = 0.235 J/g( o C ) 62

63. Practice Problem What is the final temp of silver if the temperature of a 5.8 g sample of silver starts out at 30.0 o C and 40.5 J of heat is added? The specific heat of silver is .235 J/g( o C). 63

64. Givens: m = 5.8 g T i = 30.0 o C Q = 40.5 J C = 0.235 T f = ??? Q = mC ∆T 40.5=5.8(0.235)( T f -30.0) 40.5=1.363(T f -30.0) 40.5=1.363T f – 40.89 81.39=1.363Tf T f = 59.7139 T f = 60. o C 64

65. What quantity of heat is required to raise the temperature of 100 mL of water from 45.6 C to 52.8C? The specific heat of water is 4.184 J/g(C) and water has a density of 1.00 grams/mL. Practice Problem 65

66. Givens: Q = ? V = 100. mL T i = 45.6 o C T f = 52.8 o C C = 4.184 d water =1.00 g/ mL 100 mL = 100 g Q = mC ∆T Q=100(4.184)(52.8-45.6) Q=3012.48 Q=3010 J 66

67. Heat Transfer A 32-g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon. C p = 240 J/kg· °C ? GIVEN: m = 32 g T i = 60°C T f = 20°C Q = ? C p = 240 J/kg· °C WORK: Q = m·  T ·C p m = 32 g = 0.032 kg T = 20°C - 60°C = – 40°C Q = (0.032kg)(-40°C)(240 J/kg· °C ) Q = – 301 J 67

68. Heat Transfer How much heat is required to warm 230 g of water from 12°C to 90°C? C p = 4184 J/kg· °C GIVEN: m = 230 g T i = 12°C T f = 90°C Q = ? C p = 4184 J/kg· °C WORK: Q = m·  T ·C p m = 230 g = 0.23 kg T = 90°C - 12°C = 78°C Q = (0.23kg)(78°C)(4184 J/kg· °C ) Q = 75,061 J 68

69. Specific Heat 2. Some things heat up or cool down faster than others. Land heats up and cools down faster than water 69

70. b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). 1) C water = 4184 J / kg C 2) C sand = 664 J / kg C This is why land heats up quickly during the day and cools quickly at night and why water takes longer. 70

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72. Heat Transfer from object to surroundings The heat transfer from an object to its surroundings and from the surroundings to an object is equal. One will experience heat loss, one will experience heat gain. H loss = H gain m· T ·C p (object) = Q = m· T ·C p (surroundings) 72

73. Exit Slip A wooden spoon has a mass of 20 kg and a specific heat of 1,700 J/(kg·C). Find the heat change of the block as it warms from 15 °C to 25°C. 73

74. The Transfer of Heat 74

75. Conductors and Insulators Materials are either conductors or insulators. A conductor transfers thermal energy Ex:metals-silver and steel, tile floors takes heat away from your An insulator does not transfer thermal energy well. Ex: wood, wool, straw, paper 75

76. Thermodynamics Study of the relationships between thermal energy, heat and work Heat and work increase thermal energy Heat – warming hands by a fire Work – warming hands by rubbing them together 76

77. The Zeroth Law of Thermodynamics If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other 77

78. The Zeroth Law of Thermodynamics If C is initially in thermal equilibrium with both A and B , then A and B are in thermal equilibrium with each other. 78

79. Linear Thermal Expansion Increasing the temperature of a rod causes it to expand. For moderate changes in temperature, the change in length is given by: 79

80. 1) A metal rod is 2 meters long at 20°C. If its coefficient of linear expansion is 1.2 × 10⁻⁵ °C⁻¹, find its length when heated to 50°C. 80

81. Answer ΔL = L₀αΔT = (2)(1.2 × 10⁻⁵)(50 - 20) = 0.00072 m = 0.72 mm. New length = 2.00072 m. 81

82. 2) A glass rod expands by 0.3 mm when heated from 25°C to 75°C. If its original length is 1.5 m, find the coefficient of linear expansion. 82

83. Answer α = ΔL / (L₀ΔT) = 0.0003 / (1.5 × 50) = 4 × 10⁻⁶ °C⁻¹. 83

84. 3) A steel rod has a coefficient of linear expansion of 1.1 × 10⁻⁵ °C⁻¹. If it is 3 m long at 20°C, what is the change in length when heated to 80°C? 84

85. Answer ΔL = (3)(1.1 × 10⁻⁵)(80 - 20) = 0.00198 m = 1.98 mm. 85

86. 4) If a brass rod expands by 2 mm when heated by 100°C, what would be its expansion if the temperature change were only 50°C? 86

87. Answer Since expansion is proportional to temperature change: New expansion = (50/100) × 2 = 1 mm. 87

88. 5) A railway track is made of steel rails, each 12 m long at 15°C. If the temperature rises to 45°C, and the coefficient of linear expansion for steel is 1.2 × 10⁻⁵ °C⁻¹, find the expansion per rail. 88

89. Answer ΔL = (12)(1.2 × 10⁻⁵)(45 - 15) = 4.32 mm 89

90. 6) A rod expands from 1.2 m to 1.202 m when the temperature increases from 25°C to 125°C. Find the coefficient of linear expansion. 90

91. Answer α = (0.002) / (1.2 × 100) = 1.67 × 10⁻⁵ °C⁻¹ 91

92. 7) A rod expands from 100 cm to 100.5 cm when heated from 20°C to 120°C. Find the coefficient of linear expansion. 92

93. Answer α = (0.5) / (100 × 100) = 5 × 10⁻⁶ °C⁻¹ 93

94. 9) A bridge made of steel is 500 m long at 10°C. How much will it expand at 40°C if the coefficient of linear expansion for steel is 1.2 × 10⁻⁵ °C⁻¹? 94

95. Answer ΔL = (500)(1.2 × 10⁻⁵)(40 - 10) = 18 cm 95

96. 10) A rod expands by 0.002 m when heated by 100°C. If its original length was 2 m, find its coefficient of linear expansion. 96

97. Answer α = (0.002) / (2 × 100) = 1 × 10⁻⁵ °C⁻¹. 97

98. 12) A rod of unknown material expands by 0.01 m when its temperature increases by 150°C. If its original length was 4 m, determine its coefficient of linear expansion. 98

99. Answer α = (0.01) / (4 × 150) = 1.67 × 10⁻⁵ °C⁻¹. 99

100. 13) A copper rod of length 2.5 m is heated from 30°C to 200°C. If the coefficient of linear expansion of copper is 1.7 × 10⁻⁵ °C⁻¹, calculate the final length of the rod. 100

101. Answer ΔL = (2.5)(1.7 × 10⁻⁵)(200 - 30) = 7.225 mm. Final length = 2.507225 m 101

102. 14) A laboratory experiment determines that a metal rod expands by 0.004 m when heated from 25°C to 225°C. The original length was 5 m. Identify the metal using the given table of expansion coefficients. 102

103. Answer α = (0.004) / (5 × 200) = 4 × 10⁻⁶ °C⁻¹. This does not match the listed metals, so it is likely an unknown material. 103

104. 15) A steel pipeline of length 800 m is installed at 15°C. If it is expected to reach a maximum of 45°C, calculate how much expansion should be allowed in the design (coefficient of expansion for steel: 1.2 × 10⁻⁵ °C⁻¹). 104

105. Answer ΔL = (800)(1.2 × 10⁻⁵)(45 - 15) = 28.8 cm 105

106. Applications of thermal expansion 106

107. Figure 17.3a 107

108. Figure 17.3b 108

109. Figure 17.3b 109

110. Figure 17.3c 110

111. C) A nother application of bimetallic strip Fire alarm 111

112. Example of Thermal Expansion This railroad track has a gap between segments to allow for thermal expansion. On hot days, the segments expand and fill in the gap. If there were no gaps, the track could buckle under very hot conditions. 112

113. 8) Why do gaps need to be left between railway tracks? 113

114. Answer To allow for the expansion of metal due to temperature increases. Without these gaps, the rails could bend or break due to thermal expansion. 114

115. 11) A metal wire is stretched tightly between two poles. Explain why it may sag in the summer and become tight in the winter? 115

116. Answer In summer, the wire expands and sags. In winter, it contracts and becomes tighter. 116

117. Bridges due to thermal expansion 117 On a hot day concrete bridges expand To solve this problem, we leave small gab at one end and support the other end with rollers.

118. Concrete due to thermal expansion 118 On a hot day concrete runway sections in airport expands and this cause cracking To solve this problem we leave small gabs between sections.

119. What is meant by a thermostat? 119 A thermostat is temperature operated electrical switch that uses the expansion properties of a bimetal strip.

120. Some applications to thermostat in industry 120 electric irons fish tanks home heating/cooling systems Ovens, refrigerators,  fire alarms, car flashers

121. Thermal expansion of solids, liquids and gases 121 The particles in gases and liquids move more freely than in solids, so their volume increases when heated.

122. Summary 122 Matter expands when heated and contracts when cooled Liquids expand and contract more than solids Gases expand and contract more than liquids

123. Molecular Basis for Thermal Expansion (1 of 2) We can understand linear expansion if we model the atoms as being held together by springs. When the temperature increases, the average distance between atoms also increases. As the atoms get farther apart, every dimension increases. 123

124. Molecular Basis for Thermal Expansion (2 of 2) A graph of the “spring” potential energy versus distance between neighboring atoms is not symmetrical. As the energy increases and the atoms oscillate with greater amplitude, the average distance increases. 124

125. Expanding Holes and Volume Expansion If an object has a hole in it, the hole also expands with the object, as shown. The hole does not shrink . The change in volume due to thermal expansion is given by where β is the coefficient of volume expansion and is equal to 3 α . Video Tutor Solution: Example 17.3 125

126. Table 17.1 Coefficients of Linear Expansion Material alpha in units per kelvin or degree Celsius Aluminum 2.4 times 10 to the negative fifth Brass 2.0 times 10 to the negative fifth Copper 1.7 times 10 to the negative fifth Glass 0.4 to 0.9 times 10 to the negative fifth Invar (nickel–iron alloy) 0.09 times 10 to the negative fifth Quartz (fused) 0.04 times 10 to the negative fifth Steel 1.2 times 10 to the negative fifth 126

127. Table 17.2 Coefficients of Volume Expansion Solids beta in units per kelvin or degree Celsius Aluminum 7.2 times 10 to the negative fifth Brass 6.0 times 10 to the negative fifth Copper 5.1 times 10 to the negative fifth Glass 1.2 to 2.7 times 10 to the negative fifth Invar 0.27 times 10 to the negative fifth Quartz (fused) 0.12 times 10 to the negative fifth Steel 3.6 times 10 to the negative fifth 127

128. Thermal Expansion of Water Between 0 degrees Celsius and 4 degrees Celsius, water decreases in volume with increasing temperature. Because of this anomalous behavior, lakes freeze from the top down instead of from the bottom up. 128

129. Quantity of Heat (1 of 2) Sir James Joule (1818–1889) studied how water can be warmed by vigorous stirring with a paddle wheel. 129

130. Quantity of Heat (2 of 2) The same temperature change caused by stirring can also be caused by putting the water in contact with some hotter object. The calorie (abbreviated cal) is the amount of heat required to raise the temperature of 1 gram of water from 14.5 degrees Celsius to 15.5 degrees Celsius. 130

131. Specific Heat The quantity of heat Q required to increase the temperature of a mass m of a certain material by Δ T is: The specific heat c has different values for different materials. The specific heat of water is approximately 131

132. Phase Changes The phases (or states) of matter are solid, liquid, and gas. A phase change is a transition from one phase to another. The temperature does not change during a phase change. The latent heat , L , is the heat per unit mass that is transferred in a phase change. 132

133. Heat Added to Ice at a Constant Rate Video Tutor Solution: Example 17.8 133

134. Heat of Fusion The metal gallium, shown here melting in a person’s hand, is one of the few elements that melts at room temperature. Its melting temperature is 29.8 degrees Celsius, and its heat of fusion is 134

135. Heat of Vaporization The water may be warm and it may be a hot day, but these children will feel cold when they first step out of the swimming pool. That’s because as water evaporates from their skin, it removes the heat of vaporization from their bodies. To stay warm, they will need to dry off immediately. 135

136. Conduction of Heat In conduction, heat flows from a higher to a lower temperature. Consider a solid rod of conducting material with cross-sectional area A and length L . The left end of the rod is kept at a temperature T H and the right end at a lower temperature T C . The rate that heat is transferred is: 136

137. Thermal Conductivities of Some Common Substances Substance k in watts per meter kelvin Silver 406 Copper 385 Aluminum 205 Wood 0.12 – 0.04 Concrete 0.8 Fiberglass 0.04 Styrofoam 0.027 137

138. Energy Conservation in Thermodynamic Processes 1. Introduction to Thermodynamics Thermodynamics is the branch of physics that deals with heat, work, and energy. The fundamental concept in thermodynamics is the conservation of energy , which states that: Energy cannot be created or destroyed, only transferred or converted from one form to another. 138

139. 2. Principle of Energy Conservation in Thermodynamic Processes Definition The principle of energy conservation states that in any thermodynamic process, the total energy remains constant . Energy may be transferred as heat (Q) or work (W) but the internal energy (U) of the system changes accordingly. 139

140. Examples of Energy Conservation in Thermodynamics Boiling Water: Heat from a stove transfers energy to the water, increasing its temperature . Car Engine: Chemical energy in fuel is converted into heat and mechanical work. Air Conditioners: Electrical energy is used to remove heat from a room, keeping it cool . 140

141. 3. First Law of Thermodynamics Mathematical Formula : ΔU=Q −W Where : ΔU = Change in internal energy (J) Q = Heat added to the system (J) W = Work done by the system (J) 141

142. Explanation If heat is added to a system ( +Q ), the internal energy increases . If heat is removed from a system ( -Q ), the internal energy decreases . If work is done on the system ( -W ), the internal energy increases . If work is done by the system on the surroundings ( +W ), the internal energy decreases . 142

143. Example Situations : Expanding Gas: When a gas expands in a piston, work is done by it on its surroundings, decreasing internal energy . Compressing Gas: When a gas is compressed , work is done on it, increasing internal energy. 143

144. 4. Work and Heat Transfer in Thermodynamics Work (W) Work is done by a system when it expands (gas pushing a piston). Work is done on a system when it is compressed. Measured in Joules (J). Heat (Q) Heat is energy transferred due to temperature difference. Flows from hot to cold. Measured in Joules (J). 144

145. 5. Practice Questions A. Multiple-Choice Questions 1. What happens to the internal energy of a gas when work is done on it? A) It decreases B) It increases C) It remains constant D) It depends on the heat transfer 145

146. 1. What happens to the internal energy of a gas when work is done on it? A) It decreases B) It increases ✅ C) It remains constant D) It depends on the heat transfer 146

147. 2. Which equation represents the first law of thermodynamics? A ) W=Q− U B) ΔU=Q+W C) ΔU=Q− W D) Q=W− ΔU 147

148. 2. Which equation represents the first law of thermodynamics? A ) W=Q− U B) ΔU=Q+W C) ΔU=Q− W ✅ D) Q=W− ΔU 148

149. 3. In a heat engine, what happens to some of the input energy? A ) It is destroyed B) It is converted into matter C) It is converted into useful work and heat loss D) It disappears 149

150. 3. In a heat engine, what happens to some of the input energy? A ) It is destroyed B) It is converted into matter C) It is converted into useful work and heat loss ✅ D) It disappears 150

151. 4. If a system absorbs 100J of heat and does 40J of work, what is the change in internal energy? A ) 60J B) 140J C) -60J D) 40J 151

152. 4. If a system absorbs 100J of heat and does 40J of work, what is the change in internal energy? A ) 60J ✅ B) 140J C) -60J D) 40J 152

153. 5. What does the first law of thermodynamics primarily describe? A ) Conservation of energy B) The expansion of gases C) The cooling of objects D) The properties of solids 153

154. 5. What does the first law of thermodynamics primarily describe? A ) Conservation of energy ✅ B) The expansion of gases C) The cooling of objects D) The properties of solids 154

155. Free response questions 6. Explain why the first law of thermodynamics is also called the law of energy conservation. 155

156. 6. Explain why the first law of thermodynamics is also called the law of energy conservation. Answer : The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added minus the work done. This aligns with the principle of conservation of energy, which states that energy cannot be created or destroyed, only transformed. 156

157. 7. A system receives 500J of heat and performs 200J of work. Calculate the change in internal energy. 157

158. 7. A system receives 500J of heat and performs 200J of work. Calculate the change in internal energy. Answer : ΔU=Q −W=500J− 200J=300J. 158

159. 8. Describe a real-life example of energy conservation in a thermodynamic process. 159

160. 8. Describe a real-life example of energy conservation in a thermodynamic process. Answer : A steam engine converts heat energy from burning fuel into mechanical work, demonstrating the conservation of energy. 160

161. 9. How does heat transfer occur in a thermodynamic system? 161

162. 9. How does heat transfer occur in a thermodynamic system? Answer : Heat transfer occurs through conduction, convection, or radiation, depending on the medium and temperature difference. 162

163. 10. What is the relationship between work and heat in thermodynamic processes? 163

164. 10. What is the relationship between work and heat in thermodynamic processes? Answer : Heat and work are the two ways energy is transferred in thermodynamic systems. They influence the internal energy of a system as described by the first law of thermodynamics. 164

165. 6. Summary Energy conservation means that energy cannot be created or destroyed. The first law of thermodynamics states that the internal energy change is equal to heat added minus work done. Work (W) is energy used to move objects or expand gases. Heat (Q) is energy transfer due to temperature differences. Applications include engines, refrigerators, and heat pumps. 165