Lecture 1 Fall 24-25_2 PDF

Chapter 18:Temperature, heat and the first law of thermodynamics (Fundamental of Physics, 10th edition) ❑ 2 lectures (one week, 4 hours) for this chapter: Lecture 1: Heat, temperature, zeroth law of thermodynamics, absorption of heat: heat capacity, specific heat capacity, m...

Chapter 18:Temperature, heat and the first law of thermodynamics (Fundamental of Physics, 10th edition) ❑ 2 lectures (one week, 4 hours) for this chapter: Lecture 1: Heat, temperature, zeroth law of thermodynamics, absorption of heat: heat capacity, specific heat capacity, molar specific heat, latent heat of fusion and vaporization, related problems. Lecture 2: Concept of work done in thermodynamics, First law of thermodynamics and its applications, related problems.  Lecture 1 18.1 Thermodynamics : One of the principal branches of physics and engineering is thermodynamics, which is the study and application of the thermal energy (often called the internal energy) of systems. One of the central concepts of thermodynamics is temperature. Temperature : Temperature is an SI base quantity related to our sense of hot and cold. It is measured with a thermometer, which contains a working substance with a measurable property, such as length or pressure, that changes in a regular way as the substance becomes hotter or colder.  Recall: Temperature Scales Here T= temperature in Kelvin scale 𝑇𝐶 = temperature in Celsius scale 𝑇𝐹 = temperature in Fahrenheit scale Class Work: Temperature difference ▪ The temperature of an object is 25 °C. After absorbing some amount of heat the temperature of the object becomes 55 °C. What would be the temperature difference in Kelvin scale, in this case? Solve it by yourself! The Zeroth Law of Thermodynamics: Suppose that, as in following Fig (a) , we put a thermoscope (which we shall call body T) into intimate contact with another body (body A).The entire system is confined within a thick-walled insulating box. “If bodies A and B are each in thermal equilibrium with a third body T, then A and B are in thermal equilibrium with each other.” In less formal language, the message of the zeroth law is: “Every body has a property called temperature. When two bodies are in thermal equilibrium, their temperatures are equal. And vice versa.” 18.4 ABSORPTION OF HEAT : Temperature and Heat A change in temperature is due to a change in the thermal energy of the system because of a transfer of energy between the system and the system’s environment. The transferred energy is called ‘ HEAT’ and is symbolized Q. Heat is positive when energy is transferred to a system’s thermal energy from its environment (we say that heat is absorbed by the system). Heat is negative when energy is transferred from a system’s thermal energy to its environment (we say that heat is released or lost by the system). Heat: Heat is the energy transferred between a system and its environment because of a temperature difference that exists between them. 1 cal = 3.968 × 10-3 Btu = 4.1868 J. The Absorption of Heat by Solids and Liquids Heat Capacity The heat capacity, C of an object is the proportionality constant between the heat Q that the object absorbs or loses and the resulting temperature change ΔT of the object; that is, Q ∝ ΔT Q = C ΔT = C ( Tf – Ti ) C = Q/ ΔT in which Ti and Tf are the initial and final temperatures of the object, respectively. Unit : cal/ C° or J/K Specific Heat Capacity Two objects made of the same material—say, marble—will have heat capacities proportional to their masses. It is therefore convenient to define a “heat capacity per unit mass” or specific heat c that refers not to an object but to a unit mass of the material of which the object is made. [c = C/m C = mc] Q = C ΔT = mcΔT c = Q/mΔT Unit : cal/g-C° or J/kg-K  Molar Specific Heat In many instances the most convenient unit for specifying the amount of a substance is the mole (mol), where 1 mole = NA= 6.02 × 1023 elementary units (atoms or molecule) of any substance. The molar specific heat of a material is the heat capacity per mole, which means per 6.02 × 1023 elementary units of the material. Thus 1 mol of aluminum means 6.02x1023 atoms (the atom is the elementary unit), and 1 mol of aluminum oxide means 6.02 x1023 molecules (the molecule is the elementary unit of the compound). When quantities are expressed in moles, specific heats must also involve moles (rather than a mass unit); they are then called molar specific heats. The molar specific heat, cm of a material is the heat capacity per mole. Q = ncm ΔT cm = Q/nΔT Unit : cal/mol-C° or J/ mol-K Problem 23 : A small electric immersion heater is used to heat 100 g of water for a cup of instant coffee. The heater is labeled “200 watts” (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from 23° C to 100° C, ignoring any heat losses. Solution: m = 0.100 kg P = 200 W c = 4190 J/kg-K Ti = 23°C = 23+273 = 296 K Tf = 100°C = 100 +273 K= 373 K ∆T = Tf –Ti = 373 – 296 K=77 K Q = cm ∆T P = W/t P = Q/t 𝑸 𝒎𝒄∆T 𝟎.𝟏𝟎𝟎( 𝟒𝟏𝟗𝟎)(𝟕𝟕) Now, t = = = = 160 sec (Ans) 𝑷 𝑷 𝟐𝟎𝟎 Problem 24 : A certain substance has a mass per mole of 50.0 g/mol. When 314 J is added as heat to a 30.0 g sample, the sample’s temperature rises from 25.0° C to 45.0° C. What are the (a) specific heat and (b) molar specific heat of this substance? (c) How many moles are in the sample? Solution : Given, m = 30 g = 30 × 10 -3 kg ; Tf = 45°C ; Ti = 25°C ; Q = 314 J ∆T = (45 + 273)K – (25+273)K = (45 -25 )K = 20 K (a) Q = mc∆T 𝑄 314 c= = = 523 J/kg-K 𝑚 ∆T 30 × 10 −3 ×20 (c) Mass of sample, m = Msam = 30 g = 30 × 10 -3 kg And [ Mass per mole, M = 50 g =50 × 10 -3 kg ] So, Msam = nM 𝑀𝑠𝑎𝑚 30 × 10 −3 Now, number of mole, n = = = 0.600 mol 𝑀 50 × 10 −3 (b) Q = ncm∆T 𝑄 314 cm = = = 26.2 J/mol-K 𝑚 ∆T 0.600 ×20  18.4: Heats of transformation ❑ Specific Heat: Heat required to change the temperature of the material or sample. Q = cm ∆T ❑ Latent Heat: Heat required to change the state of the material. Latent Heat of Fusion Latent Heat of Vaporization Solid -------------------- Liquid --------------------Vapor Liquid Vapor-------------------Liquid Liquid------------------- Solid Q = mL When energy is absorbed as heat by a solid or liquid, the temperature of the sample does not necessarily rise. Instead, the sample may change from one phase (or state) to another. The amount of energy required per unit mass to change the state (at constant temperature) of a particular material is its heat of transformation L. Thus, when a sample of mass m completely undergoes a phase change, the total energy transferred is Q = mL ❑ Heat of Vaporization The heat of vaporization LV is the amount of energy per unit mass that must be added to vaporize a liquid or that must be removed to condense a gas. For water at its normal boiling or condensation temperature, Q = m Lv [LV = 539 cal/g = 40.7 kJ/mol = 2256 kJ/kg] ❑ Heat of Fusion The heat of fusion Lf is the amount of energy per unit mass that must be added to melt a solid or that must be removed to freeze a liquid. For water at its normal freezing or melting temperature, Q = m Lf [Lf = 79.5 cal/g = 6.01 kJ/mol = 333 kJ/kg] Problem 27 : Calculate the minimum amount of energy, in joules, required to completely melt 130 g of silver initially at 15.0° C. Consider melting temperature of silver is 1235 K and its specific heat capacity and latent heat of fusion are 236 J/kg-K and 105 KJ/kg, respectively. Solution: Here, c = 236 J/kg-K, m = 130 g = 0.130 kg Ti =15.0 °C = (273+15)K = 288 K, Tf = 1235 K ∆𝑇 = (1235 − 288)𝐾 = 947 K Now, specific heat Q1 = cm∆𝑻 = 236(0.130)947 = 2.91×104 J Again, for the state change [ Solid to Liquid] Latent heat of fusion, Q2 = mLf = (0.130) (105 × 103) = 1.36×104 J [Lf = 105×103 J/kg] The total heat required, Q = Q1 +Q2 = (2.91×104 + 1.36×104 ) J = 4.27×104 J Problem 28 : How much water remains unfrozen after 50.2 kJ is transferred as heat from 260 g of liquid water initially at its freezing point? Solution: Q = 50.2 kJ = 50.2 × 103 J Lf = 333 kJ/kg = 333 × 103 J/kg Mass of frozen water (ice), m =? Q = mLf 𝑄 50.2×103 m= = = 0.151 kg Lf 333×103 Liquid water, m1 = 260 g = 0.260 kg The amount of water that remains unfrozen (liquid water) = m1 – m = (0.260 – 0.151) kg = 0.109 kg = 109 g Problems for Practice: (Chapter 18, Fundamentals of Physics, 10th Edition) ❑ Absorption of Heat: 22, 25, 26, 31,36