Thermodynamics Element chemistry and Heat capacity

Chemistry

1. Elementary Chemistry and Thermodynamics I (CHEM8010) The School of Workforce Development, Continuing Education & Online Learning Week 4 Thermodynamics (cont’d and heat capacity)

2. Recap Inter and intra-molecular forces Laws of Thermodynamics Zeroth and first laws of thermodynamics Temperature, units of temperature and measuring methods Pressure, units and some devices for measuring pressure

3. Laws of thermodynamics, zeroth and first laws Second law of thermodynamics, entropy Third law of thermodynamics Internal energy, work and heat Heat capacity and specific heat capacity Lecture 4 AGEN DA

4. Zeroth law of thermodynamics If two bodies are in thermal equilibrium with a third body, they are also in equilibrium with each other. As two mechanical system in equilibrium have the same pressure , two systems in thermal equilibrium with each other have the same temperature. The zeroth law of thermodynamics is the basis for definition of temperature. Development of a thermal equilibrium in a closed system over time through a  heat flow  that levels out temperature differences Thermal equilibrium - Wikipedia

5. First Law of Thermodynamics Energy can be converted from one form to another, but it is neither created nor destroyed. Energy can be transferred between the system and surroundings. Total energy in an isolated system remains constant.

6. Second Law of Thermodynamics The first thermodynamics law allows energy transfer from one body to another and states that energy cannot be destroyed or created, but it does not predict the direction of the energy. The second law of thermodynamics specifies that heat flows from hot body to cold body, and the other way around is impossible, unless some work is done- e .g. in refrigerator, energy or work is given to the system to transfer heat from cold environment (inside refrigerator) to warm environment (room temperature). Direction of heat is from hot to cold

7. Other statements of the second law are: If heat is given from a heat reservoir to a system (e. g. an engine or a boiler), some of this energy or heat can be converted to work (at engine or turbine), but some of the heat will be given to cold reservoir (e. g. at cooling system or condenser). The second law implies that no machine can be built that works with 100% efficiency. Second Law of Thermodynamics (cont’d)

8. The amount of energy that is not available for work is called entropy . Entropy is mathematically expressed as: Where Δ S denotes entropy change, Q is the heat transfer during a process and T is the absolute temperature at which process occurs. The SI unit for entropy is  joules per kelvin (J⋅K −1 ) or kg⋅m 2 ⋅s −2 ⋅K −1 Entropy is an extensive function (depends on the amount of matter in the system)   Second Law of Thermodynamics (cont’d)

9. Entropy is a measure of disorder in the system, the higher the entropy, the higher disorder. The second law states that entropy of the universe is always increasing Δs universe or also called Δs total = Δs system + Δs surrounding > 0 Note: Δs system or Δs surrounding each can be positive or negative, but the sum (total ) of the two is always positive. Δs system is a state function (only depends on the initial and final state and independent of how system reached the final state. Second Law of Thermodynamics (cont’d)

10. At zero temperature the system must be in a state with the minimum thermal energy. This statement holds true if the perfect crystal has only one state with minimum energy. Entropy is related to the number of possible microstates. S = k B lnW where k B is the Boltzmann constant, and W is the number of microstates. Single possible configuration for a system at absolute zero, i.e., only one microstate is accessible. Third Law of Thermodynamics

11. Third Law of Thermodynamics The  entropy   S of a system approaches a constant value when its temperature approaches absolute zero . In the other words, The entropy change; ΔS will go to zero when T goes to : Another statement of third law is about unattainability of absolute zero. “It is impossible to achieve a temperature of absolute zero.”

12. Laws of Thermodynamics (review) The 4 Laws of Thermodynamics - YouTube

13. Internal Energy U Recall: First law of thermodynamics introduces internal energy U for the system. It is related to molecular motions and intermolecular interaction First law of thermodynamics in mathematical form is defined as: ΔU= U final - U initial = Q + W (closed system at rest) ΔU is the change in internal energy, Q is the heat flow into the system from surrounding, W is the work done by the surroundings on the system. U final is the final state and U initia l is the initial state of the system

14. The molecular motions of molecules comprised of translational, rotational and vibrational motions. The kinetic energy of molecules can be written as: K E =1/2 mV 2 where m is the molecular mass and v is velocity of molecules There is also a potential energy between molecule. This is due to the cohesive force that tends to hold molecules together in a distance The cohesive forces is stronger in solids and liquids compared to gasses. The sum of the kinetic and potential energies within a body or substance is known as the internal energy of the substance. Internal Energy U (cont’d)

15. The total molar internal energy of a substance is expressed as: Where U m is total molar internal energy of the substance, U tr,m is molar translational, U rot,m is molar rotational, U vib.m is molar vibrational, U el , m is molar electronic, Ui ntermo , m is molar intermolecular and U rest , m is molar rest mass of electrons and nuclei. Increasing temperature increase molecular motions and thus the internal energy of the substance. Internal Energy U (cont’d)

16. Internal Energy U (cont’d) What is Internal Energy in Physics? Thermal Physics - YouTube

17. Work The most common form of work done on a system is called P-V work. The P-V work in a thermodynamic system is a work that causes a change in the volume of the system e. g. a contraction or expansion of the system against an external pressure. Assume a system confined by a piston with an external constant pressure P. When the gas expands, it pushes the piston up. Expansion and contraction under constant external pressure. During expansion work is done by the system . During contraction work is done on the system

18. The work done by the gas on the piston is calculated as: W= P Δ V = P(V 2 -V 1 ) This is a common process when a substance in one of the physical state (liquid, solid or gas) is expanding or contracting under atmospheric pressure . If the volume of the system do not change during the process, w=0 (as Δ V=0) Expansion and contraction under constant external pressure. During expansion work is done by the system . During contraction work is done on the system Work

19. Practice problem A gas is heated slowly and expands at a constant pressure of 3.67 x 10 4 Pa from a volume of 385 cm 3 to 875 cm 3 . Find w in joules. Solution: W = - P ext (V 2 –V 1 ) ΔV = V 2 – V 1 = 875 cm 3 - 385 cm 3 = 490 cm 3 = 4.9 x 10 -4 m 3 W= - 3.67 x10 4 N/m 2 x 4.9 x 10 -4 m 3 = - 18.0 J

20. Heat Heat can be referred to as an energy transfer process that may be transferred from one body to another, due to a difference in the temperature of the bodies. when heat supplied to a body, will increase the internal energy of the body. Conversely, if heat is removed from the body the internal energy will decrease. ΔU = Q + W (q > 0 giving heat to system Δ U > 0 internal energy increases) (q < 0 giving heat to surroundings Δ U < 0 internal energy decreases )

21. Heat (cont’d) Adding heat increases both the kinetic and potential energy within the body. The velocity of molecules increases, thus increasing the kinetic energy. At the same time, adding heat to the body causes it to expand, which forces the molecules further apart against their cohesive force, thus increasing the potential energy . Removing heat decreases both the kinetic and potential energy within the body. The velocity of molecules decreases, thus decreasing the kinetic energy . At the same time, removing heat from the body causes it to contract, which moves the molecules closer together under their cohesive force, thus decreasing the potential energy.

22. Heat capacity Recall: internal energy is an extensive property; depends on the amount of matter. Heat given to a system increases internal energy (U), so it may increase temperature (T) The ratio of the heat (q) added to or removed from a substance to the change in temperature (∆T) produced is called the heat capacity (C) of the substance. From definition, the unit of heat capacity should be energy/(temperature); J/K or J/ °C  

23. Specific Heat Specific heat or specific heat capacity is defined as: “The quantity of heat required to change the temperature of a unit mass of the substance by one degree”. Specific heat can be described as the heat needed to be transferred into or out of the substance to change its temperature by 1 °C (or 1 K). Mathematically: From definition, the unit of specific heat should be energy/( mass.temperature ). The unit of specific heat is J/( g.°C ) or kJ/( kg°C ).  

24. Specific Heat Examples: the specific heat of water is given as 4.2 kJ/kg K. This means that 4.2 kJ of heat must be transferred to 1 kg of water in order to raise its temperature by 1 K (or by 1°C). Likewise, to cool 1 kg of water by 1 K would require that 1 kJ of heat be removed from the water. The specific heat is temperature dependent, however, for practical cases, the specific heat of liquid water can be assumed to be constant . Specific heat of some common substances

25. Learning check Lead has specific heat of 128J/Kg ° C, water has 4187 J/ Kg ° C . Which will change temperature more easily? Water Lead

26. Determine which will require more energy? And why? 10 kg of wood (1.76 J/ g ° C ) or 20 kg of copper(0.385 J/ g ° C ) 3kg of Lead (0.129 J/ g ° C ) or 1 kg of Steel (0.466 J/ g ° C ) 8kg Aluminum (0.89 J/ g ° C ) or 2kg Water (4.184 J/ g ° C ) Answers: a) Wood b) Steel c) Water

27. Example 1) How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•°C)? Solution: The substances undergoing reaction are called  Specific Heat ∆T = 43.87 °C – 28.3 °C = 15.6 °C  

28. Example 2) A two-ton slab of cast iron, at a temperature of 14 °C, has 20 MJ of heat transferred to it. Calculate its final temperature. Specific heat for cast iron is 0.544 kJ/kg °C. Solution: Specific Heat Q T2= 32.38 °C  

29. Specific Heat Example 3) 250 kJ of heat energy is supplied to 10 kg of iron which is initially at a temperature of 15 ° C. If the specific heat capacity of iron is 500 J/( kg ° C ) determine its final temperature? Solution: 250 X 10 3 J = 10 kg X 500 J/( kg ° C ) X (T 2 – 15 ° C )  (T 2 – 15 ° C) = 50  ° C T 2 = 65 ° C  

30. Specific Heat Example 4) A 2.71g of metal is heated to 137 °C and placed into 100.0 g of water at 25.0 °C. If the final temperature of the water was 36.4 °C, what is the specific heat capacity of the metal? The specific heat of water is 4.184 J/gºC. Solution: The system is metal, and the surroundings is water. , (First Law of thermodynamic) Final temperature of system and surrounding is same.  

31. Specific Heat Example 5) 45.0 g sample A is heated to 97.2 °C and placed into 75.3 g of water originally at 32.0 °C. If the final temperature of the water was 46.2 °C, what is the specific heat capacity of the Sample A? Answer= 1.95   The substances undergoing reaction are called 

32. Example 6) A 10-cm diameter copper ball is to be heated from 100°C to an average temperature of 150°C in 30 minutes. Taking the average density and specific heat of copper in this temperature range to be ρ= 8950 kg/m3 and Cp = 0.395 kJ/kg · °C, respectively, determine (a) the total amount of heat transfer to the copper ball, (b) the average rate of heat transfer to the ball, and (c) the average heat flux. SOLUTION The copper ball is to be heated from 100°C to 150°C. The total heat transfer, the average rate of heat transfer, and the average heat flux are to be determined. Assumptions Constant properties can be used for copper at the average temperature. Properties The average density and specific heat of copper are given to be 8950 kg/m3 and Cp 0.395 kJ/kg · °C. The substances undergoing reaction are called  Specific Heat

33. Solution a) Energy transfer to the system = Energy increase of the system Q= ∆U = mC ave (T2 - T1) where m = ρ v= = (8950 kg/m3)(0.1 m) 3 = 4.69 kg Substituting, Q= (4.69 kg)(0.395 kJ/kg · °C)(150 - 100)°C = 92.6 kJ Therefore, 92.6 kJ of heat needs to be transferred to the copper ball to heat it from 100°C to 150°C.   The substances undergoing reaction are called  Specific Heat

34. Solution b) The rate of heat transfer normally changes during a process with time. However, we can determine the average rate of heat transfer by dividing the total amount of heat transfer by the time interval. Q= ∆U = mC ave (T 2 - T 1 ) where m = ρ v= = (8950 kg/m 3 )(0.1 m) 3 = 4.69 kg ave = = 0.0514 kJ/s = 51.4 W   Specific Heat

35. Solution c: The area of sphere is calculated as: A=4 П r 2 =4 П (d/2) 2 = 4 П (0.1 m/2) 2 = 0.03 m 2 Average heat flux is the average flow of energy per unit area per unit time. From section b: ave = 51.4 W Average heat flux = 51.4 W/0.03 m 2 = 1.54 W/ m 2   Specific Heat

36. Specific heat value for any substance varies with temperature and pressure. Two important conditions are when specific heat is measure under constant P or V. c P = specific heat at constant P c V = specific heat at constant V c P is always greater than c V (why?) c p and c v are slightly different for liquids and solids. it is practically easier to measure c P for liquids and solids . For liquids and solids. c P is usually used . c p and c v are significantly different for gasses . Specific Heat

37. Summary Energy can be transferred between the system and surroundings, but the total energy of the universe is constant. The first law do not specify the direction of the heat. The second law of thermodynamics specifies that heat flows from hot body to cold body. The second law implies that no machine can be built that works with 100% efficiency. The second law introduces a state function called entropy which determines the amount of energy that is not available for work. Entropy is a measure of disorder in the system. Third law concerns about unattainability of absolute zero temperature

38. Internal energy is related to different types of molecular motions, including translational, rotational and vibrational. And intermolecular energy. The most common form of work done on a system is called P-V work which is related to change in volume of the system. Heat can be referred to as an energy transfer process between two bodies with different temperatures. Specific heat is described as the heat needed to be transferred into or out of the substance to change its temperature by 1 °C. Specific heat value is related to molecular properties of substance and is different for different substances. Summary