Energy Transfer and Thermodynamics - Assignment Example

Question 1: The four thermodynamic laws include the zeroth law of thermodynamics, the first law of thermodynamics, the second law of thermodynamics and the third law of thermodynamic. Zeroth law of thermodynamics: This law is about thermo equilibrium between two or more systems in contact, thermo equilibrium is achieved when a system temperature does not change over time i.e. S1, s2, s3, then if the three systems are at equilibrium then if Temperature S1 = temperature S2, and that temperature S2 = temperature S3 then we can conclude that the temperature of S1 = temperature S3. The first law of thermodynamics: This law states that energy cannot be destroyed, however energy can be transformed from one form to another example electric energy to heat energy and therefore the energy of a system plus the surrounding is usually constant, if we assume that heat equivalent to H is added to a system and some work equivalent to w is performed by system on the surrounding then the net energy will be H – w = I, I in this case will be the internal energy of a system. Second law of thermodynamics: This law discusses the increase of entropy and that entropy never decreases, according to this law entropy of a system that is not at equilibrium increases over time, this continues until it reaches equilibrium. The third law of thermodynamics: This is a law about absolute zero, according to this law as a system approaches absolute zero its minimum entropy value is achieved and all processes cease. Entropy S = 0 K for a pure crystal. According to this theory therefore the entropy approaches absolute zero depends on the temperature. Question 2: Entropy is the measure of the disorder of a thermodynamic system. Entropy is denoted as S and refers to the energy that is unavailable to perform work. When ice melts into water the following process occurs, ice is a solid and this is because the water molecules in ice have strong hydrogen bonds and molecules do not move independently, however these water molecules in ice possess kinetic energy and the only motion is vibration. An increase in temperature signifies an increase in kinetic energy, as kinetic energy increases vibration increases and the force holding the molecules in a crystalline form is overcome, as a result the hydrogen bonds are broken and the ice melts into water molecules. We consider two systems, the ice and the surrounding room, heat energy is transferred from the warmer surrounding to the cooler system which is the ice, as a result the entropy of the surrounding system declines and that of the ice, however the decline in entropy for the surrounding room decline is less than that of the ice Question 3: The heat of formation of compounds is denoted as DHf and this heat of formation is equal to the compounds enthalpy change. The above reaction shows the formation of nitrogen peroxide through oxidation, The enthalpy change in the above reaction is -56.53 kilojoules per mole. Therefore there is a decline in enthalpy Question 4: i. Absolute zero: Kelvin- is zero on the Kelvin scale, 0K Celsius-is negative 273 degrees Celsius Fahrenheit- 459 degrees F Rankine scales- zero on the ranking scale ii. Boiling point of water 100 degrees Celsius In Kelvin If 00c = -2730K Then 1000c = -2730K + 100 = 1730K iii. Increase of 30°C during a heating process in Kelvin Assume an increase from 0°C to 30°C C= K − 273.15 0°C = -2730K 30°C=-2430K Which is similar to a rise in temperature from -2730K to -2430K? Therefore this is equivalent to 300K iv. Increase of by 60°F during a heating process R, K and °C Assume this is an increase from 0°F to 60°F a. °C 60°F F=C (1.8) + 32 60= C (1.8) + 32 28 = C (1.8) C = 15.55°C 0°F F=C (1.8) + 32 0 =C (1.8) + 32 -32 = C (1.8) C = -17.78°C 60°F Increase = -17.78 – 15.55 = 33.33°C b. R 60°F R = F + 459.67 R = 60 + 459.67 R = 519.67 Rankine scales 0°F R = F + 459.67 R = 0 + 459.67 R = 459.67 Rankine scales 60°F Increase = 519.67 – 459.67= 60 Rankine scales c. K 60°F K = (F + 459.67) / 1.8 K = (60 + 459.67) / 1.8 K = 288.71° Kelvin 0°F K = (F + 459.67) / 1.8 K = (0 + 459.67) / 1.8 K = 255.37° Kelvin 60°F Increase = 288.71– 255.37= 33.34 ° Kelvin Question 5: Equilibrium is achieved through a process called thermolization, it is achieved when a system is undisturbed, and therefore when work is performed on a system a non equilibrium state is achieved. Question 6: An equilibrium state is a state where a system’s properties do not change over time, example of equilibrium state is water in a room at room temperature and pressure. a steady state on the other hand is a condition whereby the system may change over time in one direction but this change is continually balanced by a change in another system an example of a stable state is ice melting in a room, uniform refers to a constant change in one direction, example uniform or constant acceleration in one direction. Question 7 A closed system is a system that is isolated from its environment, the system can exchange both heat and energy but cannot exchange matter with the surrounding. i) Rechargeable battery It is a closed system in that there is exchange of energy but no exchange of matter ii) Household refrigerator It is a closed system in that there is exchange of energy and heat but no exchange of matter iii) Radiator It is an open system in that there is the exchange of heat and exchange of matter which is water with the environment. Question 8 A solid has a definite shape and volume, a liquid has no definite shape but has a definite volume, for a gas it does not have a definite shape and also does not have a definite volume. Question 9 Heat transfer in thermodynamics refers to the transfer of heat from high temperature systems to lower temperature items, heat transfer occurs until thermo equilibrium is achieved, it is also evident that heat transfer cannot be stopped but the process can be slowed down. Heat transfer occurs through radiation, convection or conduction. Question 10 Enthalpy is a measure of the thermodynamic potentials of systems and is used to determine the heat transfer in systems. Internal energy on the other hand is the total energy possessed by systems, this include kinetic energy, electric energy and the potential energy of a system Question 11 Mass flow rate is 4kg/s and the heat of combustion for C3H8 is 46450kJ/kg. Heat release rate is: Heat release rate = mass flow rate X heat of combustion Heat release rate = 4kg/s X 46450kJ/kg Heat release rate = 185800 kJ/ second Question 12 Fourier’s Law: The Fourier’s Law is a description of conduction of heat taking into consideration the temperature and the distribution properties of the medium, example metal, gas and liquids. He introduced the thermo conductivity of a medium by stating that K = -q/ T where K is thermo conductivity and T is change in time. Thermal conductivity: Thermal conductivity can be defined as the measure of any materials ability to transfer heat through conduction. Thermal conductivity of metals, insulating materials and gases: Metals have the highest thermo conductivity level which is over 100 w/m-k while gases have the lowest conductivity level which is less then one but greater than 0w/m-k, insulating materials have a higher thermal conductivity level than gases but have a lower thermal conductivity level compared to metals. Fourier’s law and the minus sign: According to the Fourier’s law heat transfer occurs in the direction of lower temperature, this means that heat flows from high temperature to low temperature, therefore the Fourier’s law has a negative sign. Question 13 Stefen-Boltzman Law: Stefen Boltzman Law analyses the relationship between heat energy radiation, surface area and temperature. According to this theory the total heat energy radiated given a certain surface area of a black body is proportional to the temperature. This law also states that a portion of energy that is not radiated is characterized by emissivity. Emissivity: Emissivity is defined as the ratio of the energy radiated to the energy radiated by a black body given that the two materials are at the same temperature. View factor and the rate of heat transfer: The view factor is the proportion of radiation that is exchanged by two surface points, it is important in the study of radiation whereby radiation travels in a straight line and that in a convex surface radiation that leaves this surface will strike this surface again. Blackbody: A black body is an object that absorbs all the radiation from its surroundings, it does not reflect radiation and no radiation passes through. Question 14 Newton’s Law of Cooling: Newton Law of cooling states that the rate of heat energy transfer of a system is proportional to the difference in the level of temperature between the surrounding and the body. The law states that dQ/dT = H A(T1 – T2), where Q is the heat energy in joules, T is time, H is the heat transfer coefficient, A is the surface area, T1 is the temperature of the body and T 2 is the temperature of the surrounding. Heat transfer coefficient: The heat transfer coefficient is a measure of proportionality between the heat energy flux of a body and the flow of heat, it is determined using the formula: H = dQ/ (ATS) where H is the heat transfer coefficient, Q is the heat energy in joules, A is the surface area, dT is the temperature differences between the body and the surrounding and S is the time in seconds. Nusselt number: The Nusselt number is a ratio depicting the relationship between conductive and convection heat energy transfer, it is calculated using the formula: Nu = convectional transfer of heat / conductive transfer of heat. Two types of convection There are two types of convection and they include natural convection and forced convection. Question 15 Heat of combustion: The heat of combustion can be defined as the amount of heat energy released by a compound during combustion with oxygen. This energy is measured in kilojoules per mole of the compound. Heat release rate: Heat release rate can be defined as the rate at which heat energy is released during combustion. Heat release rate is measured in joules per seconds. Therefore heat release rate = heat energy released in joules / time in seconds Combustion reaction: A combustion reaction occurs when a compound example a hydrocarbon reacts with oxygen or oxidizing elements and the result are compounds of elements of the reactant and oxidizing elements. Example combustion of hydrogen and oxygen: Different types of combustion: Complete combustion: this is combustion whereby the reactant is completely depleted in the combustion process, example the combustion of a hydrocarbon that yields carbon dioxide and water only. Incomplete combustion: this is combustion whereby the reactant is not completely depleted in the combustion process and this is because enough oxygen is not present. Example the combustion of a hydrocarbon that yields carbon dioxide and water, carbon, carbon monoxide Rapid combustion: in rapid combustion large amounts of light and heat energy is released, because large amount of gasses are released in the process a loud noise is produced in the process. Definitions: Specific heat capacity: The specific heat is the amount of heat energy required to change the temperature of a given quantity of compound. Example the specific heat of increasing water temperature by one degree Celsius is 4186 joules per kilogram. Latent heat: Latent heat is defined as the amount of heat energy released or required in order to change compounds from one state to another example from gaseous state into liquid state or liquid state into solid state. Calorimetry: Calorimetry is a study that involves the measure of heat increase or decline in chemical reactions or physical reactions. Combustion temperature: The combustion temperature is the temperature level in a combustion process; it is determined by measuring changes in temperature before and during combustion of compounds. Chemical equilibrium: Chemical equilibrium is achieved when chemical activities net changes of reactants remain unchanged for a given period time. This means that a chemical equilibrium state is a state where the forward and reverse reactions are equal. Question 16 Given that an engine produces 150J of work from 212J of energy Efficiency = output / input Therefore in our case: Efficiency = 150 / 212 Efficiency = 0.70755 Therefore the machine efficiency is = 70.755% Question 17 Heat capacity = mass X specific heat capacity X temperature Specific heat capacity of water = 4186 joules per kg to increase temperature by 1°C Let the 150g water at 10 degree be A Let the 100g water at 85degrees be B Heat capacity of A: Heat capacity = mass X specific heat capacity X temperature Heat capacity = 0.15X 4186X 10 = 6279 Heat capacity of B: Heat capacity = 0.1 X 4186X 85 = 35581 Heat capacity for both A and B = 6279+ 35581 = 41860 Heat capacity = mass X specific heat capacity X temperature 41860= 0.25X 4186X temperature Temperature = 40 degrees Question 18 (i)Heat is transfer rate: Q/T = [H A (T1 – T 2)]/ d where H is thermo conductivity, A is surface area, D is thickness, T1 is the hot temperature and T2 is the cold temperature. H =Thermo conductivity of Styrofoam = 0.033 W/m K A = 390 cm2 also 0.039 meters squared D = 6mm also 0.006 meters T1 – T2 = 85C which is also 358.15K Q/T = [(0.033) (0.039) (358.15)]/ 0.006 Q/T = 76.823175 (ii)Time required for the coffee temperature to drop from 85C to 70C The amount of heat transferred: Q =0.6 X 4186X 15 Q = 37674 Q/T = 76.823175 T = 37674/76.823175 T = 490.399 seconds Question 19 A flame can be defined as a visible exothermic reaction that emits light. These reactions include oxidation processes that are self sustaining or combustion. There are two types of flames and they include the luminous and the non luminous flame, the non luminous flame occurs when a fluid or gas burns in excess of oxygen and therefore blue and hotter then the luminous flame, this fame assumes a laminar flow, the luminous flame is yellow in colour, produces soot due to burning a fluid in little oxygen therefore gas or fluid is not completely burnt, also the flame is not as hot as the other type of flame and that its flow is tubular. Question 20 A fluid can be defined as matter that flows when shear stress is applied; examples of fluids include liquid and gases. Viscosity on the other hand is defined as the resistance of a fluid that exists when a fluid is being deformed, example oil is considered to have higher viscosity than water. Question 21 All the fluids are compressible to some extent and this include air and water, compression refers to the changes in density of a fluid as a result of change in temperature and pressure, air is more compressible than water and this is defined by the nature of their density, water is more dense than air, this is because when air is compressed and temperature reduced it forms a liquid and therefore volume occupied is reduced. Question 22 Given that a concrete slab length is 24 m at -8 ºC and that the linear expansion coefficient of concrete is 1 X 10-5 ºC-1 What is the change in length when the temperature is 35 ºC? Change in temperature = 35-(-8) = 43 ºC The formula: Change in length = ALT Where A is the linear expansion coefficient, L is the length and T is the change in temperature, e therefore states our problem as follows: Change in length = ALT Change in length = (1 X 10-5 ºC-1) (24) (43) Change in length = 24 X 10-5 meters which is 2.4 X 10-4 meters Question 23 Work = distance X force X cosine θ Work = 5m X 30 X cosine 0 degrees Work = 150 joules Energy used is = 150 joules Question 24 An aluminium siding is 12.45 meters at -18°C. What is the length at 37°C? Change in temperature is 55°C The formula: Change in length = ALT Where A is the linear expansion coefficient which in this case is 23 x 10-6 C-1, L is the length and T is the change in temperature, we therefore state our problem as follows: Change in length = ALT Change in length = (24 x 10-6) (12.45) (55) Change in length = 16434 x 10-6 meters Change in length = 0.016434 meters The length of the aluminium siding will be 12.45 + 0.016434 = 12.466434 meters Question 25 A boundary layer is a layer of a fluid which is at the vicinity of a bordering surface, in the bordering layer changes in flow patterns of the fluid occur. The following is a diagram showing laminar and turbulent flow of a liquid in a pipe: References: Dugdale, J. (1998). Entropy and its Physical Meaning. Wiley and Sons publishing, New York. Ness, H. (1999). Understanding Thermodynamics. Prentice Hall press, New Jersey Yunus, A. and Boles, M. (2005). Thermodynamics: an Engineering Approach. 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We consider two systems, the ice and the surrounding room, heat energy is transferred from the warmer surrounding to the cooler system which is the ice, as a result the entropy of the surrounding system declines and that of the ice, however the decline in entropy for the surrounding room decline is less than that of the ice Question 3: The heat of formation of compounds is denoted as DHf and this heat of formation is equal to the compounds enthalpy change. The above reaction shows the formation of nitrogen peroxide through oxidation, The enthalpy change in the above reaction is -56.

53 kilojoules per mole. Therefore there is a decline in enthalpy Question 4: i. Absolute zero: Kelvin- is zero on the Kelvin scale, 0K Celsius-is negative 273 degrees Celsius Fahrenheit- 459 degrees F Rankine scales- zero on the ranking scale ii. Boiling point of water 100 degrees Celsius In Kelvin If 00c = -2730K Then 1000c = -2730K + 100 = 1730K iii. Increase of 30°C during a heating process in Kelvin Assume an increase from 0°C to 30°C C= K − 273.15 0°C = -2730K 30°C=-2430K Which is similar to a rise in temperature from -2730K to -2430K?

Therefore this is equivalent to 300K iv. Increase of by 60°F during a heating process R, K and °C Assume this is an increase from 0°F to 60°F a. °C 60°F F=C (1.8) + 32 60= C (1.8) + 32 28 = C (1.8) C = 15.55°C 0°F F=C (1.8) + 32 0 =C (1.8) + 32 -32 = C (1.8) C = -17.78°C 60°F Increase = -17.78 – 15.55 = 33.33°C b. R 60°F R = F + 459.67 R = 60 + 459.67 R = 519.67 Rankine scales 0°F R = F + 459.67 R = 0 + 459.67 R = 459.67 Rankine scales 60°F Increase = 519.67 – 459.67= 60 Rankine scales c.

K 60°F K = (F + 459.67) / 1.8 K = (60 + 459.67) / 1.8 K = 288.71° Kelvin 0°F K = (F + 459.67) / 1.8 K = (0 + 459.67) / 1.8 K = 255.37° Kelvin 60°F Increase = 288.71– 255.37= 33.34 ° Kelvin Question 5: Equilibrium is achieved through a process called thermolization, it is achieved when a system is undisturbed, and therefore when work is performed on a system a non equilibrium state is achieved. Question 6: An equilibrium state is a state where a system’s properties do not change over time, example of equilibrium state is water in a room at room temperature and pressure.

a steady state on the other hand is a condition whereby the system may change over time in one direction but this change is continually balanced by a change in another system an example of a stable state is ice melting in a room, uniform refers to a constant change in one direction, example uniform or constant acceleration in one direction. Question 7 A closed system is a system that is isolated from its environment, the system can exchange both heat and energy but cannot exchange matter with the surrounding. i) Rechargeable battery It is a closed system in that there is exchange of energy but no exchange of matter ii) Household refrigerator It is a closed system in that there is exchange of energy and heat but no exchange of matter iii) Radiator It is an open system in that there is the exchange of heat and exchange of matter which is water with the environment.

Question 8 A solid has a definite shape and volume, a liquid has no definite shape but has a definite volume, for a gas it does not have a definite shape and also does not have a definite volume. Question 9 Heat transfer in thermodynamics refers to the transfer of heat from high temperature systems to lower temperature items, heat transfer occurs until thermo equilibrium is achieved, it is also evident that heat transfer cannot be stopped but the process can be slowed down. Heat transfer occurs through radiation, convection or conduction.

Question 10 Enthalpy is a measure of the thermodynamic potentials of systems and is used to determine the heat transfer in systems. Internal energy on the other hand is the total energy possessed by systems, this include kinetic energy, electric energy and the potential energy of a system Question 11 Mass flow rate is 4kg/s and the heat of combustion for C3H8 is 46450kJ/kg.

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