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Revision : Thermodynamics Formula sheet Formula sheet Formula sheet Thermodynamics key facts (1/9) β’ Heat is an energy [measured in π½] which flows from high to low temperature β’ When two bodies are in thermal equilibrium they have the same temperature β’ The S.I. unit of temperature is Kelvin (πΎ). This is related to degrees Celsius β by π πΎ = π β + 273 β’ Temperature difference Ξπ is the same in both units Thermodynamics key facts (2/9) β’ Heat energy needed to raise a temperature β’ The specific heat capacity π determines the energy π needed to raise the temperature of mass π of a substance by βπ π = π π βπ β’ Units of π will be π½ ππβ1 πΎ β1 Thermodynamics key facts (3/9) β’ Heat energy needed to change phase β’ The latent heat πΏ determines the energy π needed to change the phase of a mass π π=ππΏ β’ Units of πΏ will be π½ ππβ1 - can be fusion or vaporization β’ This energy is either absorbed (solid β liquid β gas) or released (gas β liquid β solid) β’ A phase change takes place at constant temperature Thermodynamics key facts (4/9) β’ Conduction is heat energy transfer by direct molecular contact Heat transfer π + βπ Power = π Area π΄ Thickness βπ₯ βπ βπ‘ =π π΄ Ξπ Ξπ₯ π = thermal conductivity Thermodynamics key facts (5/9) β’ Convection is heat energy transfer by the bulk flow of material Thermodynamics key facts (6/9) β’ Radiation is heat energy transfer by emission of electromagnetic radiation Power = βπ βπ‘ = π π΄ π4 π = Stefan-Boltzmann constant, π΄ = surface area of emitter, π = temperature of emitter (assumes emissivity=1) Thermodynamics key facts (7/9) β’ Ideal gas law β’ 1st form : π π = π ππ΅ π β’ π = Pressure, π = Volume, π = number of molecules, ππ΅ = Boltzmannβs constant, π = temperature [in K] β’ 2nd form : π π = π π π β’ π = number of moles, π = gas constant Thermodynamics key facts (8/9) β’ Kinetic theory of ideal gas β’ Pressure is due to molecular collisions β’ Average kinetic energy of molecules depends on temperature 1 2 ππ£ 2 = 32ππ΅ π π = mass of molecule, π£ 2 = average square speed, π = temperature Thermodynamics key facts (9/9) β’ Thermal expansion β’ Materials expand due to temperature rise βπ β’ Length πΏ increases by βπΏ = Ξ± πΏ βπ where πΌ = coefficient of linear expansion β’ Volume V increases by βπ = π½ π βπ where π½ = coefficient of volume expansion Practice exam questions: Section A πΎπΈ = 12ππ£ 2 = 32ππ΅ π Option C π = π π βπ π = ππ€ππ‘ππ ππ€ππ‘ππ βππ€ππ‘ππ = 1 × 4186 × 5 = 20930 π½ π 20930 βππ΄π = = = 12 πΎ ππ΄π ππ΄π 2 × 900 Option B Practice exam questions: Section A Heat energy loss is by conduction β option B Practice exam questions: Section A ΞπΏ = πΌ Ξπ πΏ Fractional expansion is the same β option A Reflects radiation β option A Practice exam questions: Section B π = π π βπ = 2.2 × 900 × 18 = 3.6 × 104 π½ Practice exam questions: Section B Stefan-Boltzmann law: π = π π΄ π 4 Re-arranging: π΄ = π π π4 = 70 5.67×10β8 × 2800 4 = 2.0 × 10β5 π2 Practice exam questions: Section B Ideal gas law (using moles): ππ = ππ π π = 2.2 πππ‘πππ = 2.2 × 10β3 π3 π = β130 + 273 = 143 πΎ ππ π 2.9 × 8.31 × 143 6 π= = = 1.6 × 10 ππ β3 π 2.2 × 10 Practice exam questions: Section C π = ππ€ππ‘ππ ππ€ππ‘ππ βπ + πππππππ πππππππ βπ π = 0.35 × 4186 × 3.3 + 0.25 × 387 × 3.3 = 5150 π½ π = ππππ πΏπ + ππππ ππ€ππ‘ππ βπ = 5150 π½ 5150 β (0.012 × 4186 × 21.7) πΏπ = = 3.39 × 105 π½/ππ 0.012 Practice exam questions: Section C Heat loss rate = π π΄ Ξπ βπ₯ = 0.80 × 5.0 × 13 2.4×10β3 = 2.2 × 104 π Practice exam questions: Section C Atomic mass = 4.0 × 1.66 × 10β27 = 6.64 × 10β27 ππ 1 2 ππ£ πππ 2 π£πππ = 3ππ΅ π = π = 32ππ΅ π π = 127 + 273 = 400 πΎ 3 × 1.38 × 10β23 × 400 3 β1 = 1.58 × 10 π π 6.64 × 10β27 Practice exam questions: Section C π = πππ€ππ × ππππ = 2.2 × 103 × 5.9 × 60 = 7.8 × 105 π½ π = π π βπ π 7.8 × 105 π= = = 2.4 ππ π βπ 4186 × 79 Next steps β’ Make sure you are comfortable with unit conversions β’ Review the thermodynamics key facts β’ Familiarize yourself with the thermodynamics section of the formula sheet β’ Try questions from the sample exam papers on Blackboard and/or the textbook Revision : Electricity Formula sheet Formula sheet Electricity key facts (1/9) β’ Electric charge π is an intrinsic property of the particles that make up matter, and can be positive (e.g. proton) or negative (e.g. electron) β’ The S.I. unit of charge is Coulombs (πΆ) β’ The elementary charge (on a proton or electron) is ± 1.6 × 10β19 πΆ β’ Electric current πΌ is the rate of flow of charge βπ πΌ= βπ‘ πΌ is measured in Amperes (π΄) Electricity key facts (2/9) β’ Coulombβs Law gives the force felt by two charges π1 and π2 separated by distance π πΉ π1 π π1 π2 πΉ= π2 π πΉ π2 π = 9 × 109 π π2 πΆ β2 β’ Like charges repel, opposite charges attract Electricity key facts (2/9) β’ Superposition principle for Coulombβs Law : if there are multiple charges, the forces from individual charges sum like vectors +ve πΉ1 +ve +ve πΉ2 πΉπ‘ππ‘ππ = πΉ1 + πΉ2 Electricity key facts (3/9) β’ The electric field at a point is the force a unit charge (π = 1 πΆ) would experience there πΉ πΈ= π πΉ=ππΈ β’ Can be represented by electric field lines Positive charge feels force along electric field line Negative charge feels force the other way Electricity key facts (4/9) β’ The electric potential difference Ξπ [in volts] is the work needed to move unit charge (π = 1 πΆ) between 2 points Work done = Potential Energy difference = π Ξπ β’ Electric field is the potential gradient : πΈ= βπ β βπ₯ If capacitor with plate separation π· is connected to battery with potential π, then πΈ = π/π· Electricity key facts (5/9) β’ Basic circuit principles : current πΌ is driven by a potential difference π Same current flows through all components of a series circuit Same voltage is dropped over all components of a parallel circuit Electricity key facts (6/9) β’ Ohmβs Law determines the current flowing through a resistance π π πΌ= π β’ Resistance is measured in Ohms (Ξ©) π=πΌπ Electricity key facts (6/9) β’ Resistances may be combined in series or parallel π 1 π 2 π π‘ππ‘ππ = π 1 + π 2 [R increases] π 1 1 π 2 π π‘ππ‘ππ 1 1 = + π 1 π 2 [R decreases] Electricity key facts (7/9) β’ Electrical energy is dissipated as heat by a resistor β’ Electrical Power π = πΌ π = πΌ2 π = π2 [unit is W] π Electricity key facts (8/9) β’ A capacitor is a device to store charge. Its capacitance πΆ measures the amount of charge π that can be stored for given potential difference π +π βπ π πΆ= π π=πΆπ β’ Capacitance is measured in Farads (πΉ) π β’ Capacitors may be combined in series or parallel [see lectures] Electricity key facts (9/9) β’ General circuits may be analysed using Kirchoffβs rules Kirchoffβs junction rule : the sum of currents at any junction is zero πΌ1 πΌ1 + πΌ2 β πΌ3 = 0 πΌ2 πΌ3 β’ Signs are different for inward/outward current β’ This rule arises from conservation of charge Electricity key facts (9/9) β’ General circuits may be analysed using Kirchoffβs rules Kirchoffβs loop rule : the sum of voltage changes around a closed loop is zero 4Ξ© πΌ1 9π 2Ξ© πΌ2 9 β 4 πΌ1 β 2 πΌ2 = 0 β’ Battery adds potential π, resistors subtract potential πΌπ β’ This rule arises from conservation of energy Practice exam questions: Section A Coulombβs Law: πΉ = π π1 π2 π2 1 4 Double π β πΉ decreases by β option A Current is the same β π = πΌ π β smaller voltage across smaller π β option B Practice exam questions: Section A Decreases β option B Practice exam questions: Section B π = π βππ΄π΅ π 45 βππ΄π΅ = = = 3000 π β3 π 15 × 10 Practice exam questions: Section B Ohmβs Law: πΌ = π π = 110 47×103 = 2.3 × 10β3 π΄ Practice exam questions: Section B 1 π π‘ππ‘ππ 1 1 = + π 1 π 2 1 1 1 = + 45 56 π 2 π 2 = 230 πΞ© Practice exam questions: Section C πΌ1 + πΌ2 β πΌ3 = 0 10 β 6πΌ1 β 2πΌ3 = 0 5 β 3πΌ1 β πΌ3 = 0 β4πΌ2 β 14 + 6πΌ1 β 10 = 0 β2πΌ2 β 12 + 3πΌ1 = 0 Practice exam questions: Section C πππ = β10 + π6Ξ© = 2 π β π6Ξ© = 12 π π 12 πΌ1 = = =2π΄ π 6 From before: 5 β 3πΌ1 β πΌ3 = 0 πΌ3 = 5 β 3πΌ1 = 5 β 3 × 2 = β1 π΄ Practice exam questions: Section C From before:β2πΌ2 β 12 + 3πΌ1 = 0 3πΌ1 β 12 β6 πΌ2 = = = β3 π΄ 2 2 π = πΌ3 2 π = β1 2 ×2=2π Practice exam questions: Section C Combine the 2Ξ©, 4Ξ©, 6Ξ© resistors in parallel 1 1 1 1 = + + β π ππππππππ = 1.1 Ξ© π ππππππππ 2 4 6 Combine the 1Ξ©, 1.1Ξ© resistors in series π π‘ππ‘ππ = 2.1 Ξ© Ohmβs Law: πΌ = π π π‘ππ‘ππ = 6 2.1 = 2.9 π΄ Practice exam questions: Section C Voltage across parallel combination = 1.1 2.1 π 3.1 πΌ= = = 0.52 π΄ π 6 × 6 π = 3.1 π Final words β’ Thanks to all students for their efforts in the Introduction to Physics course β’ Please fill in feedback surveys! β’ Good luck in the upcoming exams!