Energy Transfer and Thermodynamics Issues - Assignment Example

Q1. a) Zeroth Law of Thermodynamics: The zeroth law is also called law of equilibrium. It states that if a body A is in thermal equilibrium with a body B and in turn body B is in thermal equilibrium with body C then body A and C are also in thermal equilibrium with each other. b) First Law of Thermodynamics: The first law of thermodynamics is also called the law of conservation of energy and it states that the energy is always conserved; it can neither be created nor destroyed though energy can be converted from one form into another. ∆E System = 0 c) Second Law of Thermodynamics: The second law of thermodynamics is also called the law of entropy. This law predicts that the degree of randomness (entropy) of any isolated system always increases with time. There are several statements which explain second law of thermodynamics. ∆S System ≥ 0 Kelvin-Planck statement: It is impossible for any heat engine to produce net work in a complete cycle if it exchanges heat only with bodies at a single fixed temperature. Thus to produce the work the cycle should exchange heat with two reservoirs which are a different temperatures. The reservoir at high temperature is the source and other one at low temperature reservoir is the sink. Clausius statement: It is impossible to construct a device which, operating in a cycle, will produce no effect other than the transfer of heat from a colder to a hotter body. Thus in order to transfer heat from low temperature reservoir to high temperature reservoir, external work must be done on the cycle. d) Third law of thermodynamics: It is a statistical law of nature regarding entropy and the impossibility of reaching absolute zero of temperature. It states that the entropy of a substance approaches zero as its temperature approaches absolute zero. Q2) Entropy can be defined as degree of randomness. It is an index of unavailability or degradation of energy. Heat always flows from hot to cold body and thus becomes degraded or less available. Thus unavailability of energy is measured by entropy. We are usually interested in changes in entropy. Entropy is a property of system. Entropy changes always occur accompany actual heat transfer but it can also change without any heat transfer. For a reversible transfer of heat, change in entropy is given by ∆S = ∑ ∂Q/T For a reversible adiabatic process, change in entropy is zero thus called isentropic process. When ice melts into water the degree of randomness of the molecule increases and hence the entropy of the system increases. Moreover it is evident from the fact that no process between two equilibrium states is possible if it would result in decease in the total entropy of a system and its surroundings. Water in vapour form has molecules moving freely. Thus it has very high entropy. But as vapour cools, it becomes liquid. The liquid water molecules still move around, but not that freely. They already lost some entropy. Further when water cools, it becomes ice. Ice molecules can no longer move freely, but can only vibrate within the ice crystals. The entropy is now very low. Q3) Apllying Hess’s law ∆HRX = ∑∆Hproduct - ∑∆Hreactant From the enthaly table we get ∆HRX = -201.44 KJ/mol Now to calculate entropy, let us apply Clausisus inequality. For reversible process ∆S = ∑ ∆HRX /T = -201.44/T The entropy can be calculated if the temerature is known. This is an reversible exothermic reaction where the heat is released to the surrounding and the heat content of the sysytem will decrease. Thus the entropy of the system will decrease and will become less random. On the basis of this we can say that the reaction will not be spontaneous. Q4) Formula for conversion of temperature scale is: TK K = (TK - 273.15) °C = [1.80 * (TK - 273.15) + 32] °F = 1.80 TK °Rankine 1) What is absolute zero on the Kelivin, Celsius, Fahrenheit and Rankine scales? 0 K = -459.4°F = -273.15 °C = 0 °Rankine 2) The boiling point of water if 100°C what is this in Kelvin? 100 °C = 373.15 K 3) The temperature of a system rises by 30°C during a heating process. Express this rise in temperature in Kelvin. Temperature rises by 30°C = Temperature rise 30 K 4) The temperature of a system rises by 60°F during a heating process. Express this rise in temperature in R, K and °C. Temperature rise by 60°F = Temperature rise by 33.33°C = Temperature rise by 33.33K = Temperature rise by 59.4 °Rankine Q5. When the system is at equilibrium, the net work done on the system remains zero. Q6. a) Equilibrium state: Example can be two metallic containers in touch with each other and are at same temperature. Thus no heat transfer and the thermal state of equilibrium is maintained. b) Steady state: A closed container filled with two non reacting substances kept at room temperature. c) Uniform state: A container filled with a non compressible fluid. Q7. State whether the following are open or closed systems, give reasons for your answer. 1) Rechargeable battery: Closed System. No mass transfer but heat transfer takes place. 2) Household refrigerator: Closed system. No mass transfer but heat transfer takes place. 3) Radiator: Open system. Both heat and mass transfer takes place. Q8. Matter has got three states. Solid, liquid and gas. These states can be differentiated on the basis of their molecular compactness. The molecules of solid are most compact one and hence have highest density. Liquid on other hand have flowing molecules and are loosely attached with each other. The randomness is more. They are relatively less dense. Gases are least dense on the account of free molecules. Figure: Comparison of solid, liquid and gas. Q9. Thermodynamics outlines the flow heat. It says that heat flows from a body at higher temperature to a body at lower temperature spontaneously. Though reverse is possible if external work is involved. More over Thermodynamics also highlights the natural law governing the conversion of heat into work. The heat and work are mutually inter-convertible but not in exact proportion. W = JQ Where J is the Joules constant = 4.186 J/Calorie Q10. Enthalpy is the total heat content of the system which takes into account even the flow work done. Whereas, internal energy of the system is due to the virtue of molecular structure of the matter. It has nothing to do with the external work done. H = U + PV Where U is the internal Energy H is the Enthalpy PV is the flow work Q11. Given data is, Mass Flow Rate: 2kg/s Heat of Combustion for C3H8: 46450kJ/kg Heat release rate is to be determined. Heat release rate = Flow rate X Heat of combustion = 2kg/s X 46450kJ/kg = 92900 kJ/s Q12. Fourier's law is the law of heat conduction and states that the time rate of heat transfer through a material is proportional to the negative temperature gradient and to the area at right angles, to that gradient, through which the heat is flowing1. Where, is the local heat flux, [W·m−2] is the material's conductivity, [W·m−1·K−1], is the temperature gradient, [K·m−1]. The negative sign signifies the flow of heat from high to low temperature. Thermal conductivity is a material property represented by notation ‘k’ and indicates material’s ability to conduct heat. High value of thermal conductivity allows heat conduction to occur readily. It can be defined by the help of Fourier’s law as the quantity of heat, ΔQ, transmitted during given time (Δt) through a thickness x, in a direction normal to a surface of area A, induced by a temperature difference ΔT. Since thermal conductivity is a material property, it varies from material to material. In general metals exhibit very high thermal conductivity followed by insulating materials which have moderately low thermal conductivity and last comes gases including air which are poor conductors and have very low value of thermal conductivity. Q13. The Stefan–Boltzmann law2 is also known as Stefan's law and it highlights the heat transfer related to black bodies. This law states that the total energy radiated per unit surface area of a black body in unit time which is know as the emissive power, is directly proportional to the fourth power of the black body's absolute temperature T. The formula for heat transfer per unit surface area of black body is: Q = σ T4 σ is the Stefan-Boltzmann Constant = 5.67 x 10-8 Watts-m-2-°K-4 Emissivity3 of any material is a measure of its ability to radiate the absorbed energy. It can be defined as the ratio of energy radiated by a particular material to energy radiated by a black body under the same conditions. It is expressed by ‘ε’ or ‘e’. Since it is a ratio, it has got no unit and is dimensionless. For a true black body, ε =1 For real objects, ε Read More