Physics 114 – Lecture 40

Physics 114 – Lecture 40 • Chapter 14Heat • Heat Flow: Spontaneously occurs from the hotter to the colder body. Thermal equilibrium • §14.1 Heat as Energy Transfer • Flow of heat – 18th century view, flow of caloric from the hotter to the colder body • 19th century – heat was viewed as being similar to work, which was equivalent to viewing heat as a form of energy • Kinetic Theory for gases: KEave = ½ mv2ave = (3/2) kT L40-s1,8

Physics 114 – Lecture 40 • Definition of calorie (cal): • Amount of heat needed to raise 1 gram of water through a rise in temperature of 10C – more specifically from 14.5 0C to 15.5 0C • 1 kcal = 1000 cal is the amount of heat required to raise 1 kg of water through a temperature rise of 10 C • 1 kcal ≡ 1 Cal =1000 cal, the Cal being the unit of energy used by nutritionists • British system of units: 1 British thermal unit (Btu) is the amount of heat required to raise 1 lb of water through a rise in temperature of 1 0F L40-s2,8

Physics 114 – Lecture 40 • Mechanical Equivalent of Heat • Count Rumford – boring cannons • Joule (~ 1850) showed that 4.186 J = 1 cal which is equivalent to 4.186 kJ = 1 kcal • Thus heat flow is a transfer of energy from one body to another • Study example 14.1 L40-s3,8

Physics 114 – Lecture 40 • §14.2 Internal Energy • The internal energy of a system is the sum of all the energy of each molecule or atom in that system • Internal energy is sometimes referred to as thermal energy • Temperature, Heat and Internal Energy • Temperature in K: a measure of KEave of the atoms and molecules in the system • Heat: transfer of energy from one body to another because of a temperature difference between those bodies • Internal Energy: sum of energy of each molecule or atom in the system L40-s4,8

Physics 114 – Lecture 40 • Internal Energy of an Ideal Gas • For a system of N atoms, the internal energy, U, is: • U = N X KEave = N(½ mv2ave) = (3/2) N kT • With N k = n R this becomes: • U = (3/2) n RT • Note that the temperature, T, must be expressed in K • When this expression is used for molecules, the rotational KE must be taken into consideration L40-s5,8

Physics 114 – Lecture 40 • §14.3 Specific Heat • Consider an amount of heat, Q, flowing into a body of mass, m, which produces a rise in temperature, ΔT. Experimentally it is observed that Q is directly proportional to m and ΔT, but that Q does depend on the composition of the body. • The specific heat of a material is defined as: • Q = m c ΔT units of c cal/(g.0C) • Note that, from the definition of the calorie, that the above definition also defines c for water to be cwater = 1 cal/(g.0C) L40-s6,8

Physics 114 – Lecture 40 • Study examples 14.2 and 14.3 • §14.4 Calorimetry – Solving Problems • We need to exercise care in describing various types of system, which is the collection of bodies under consideration. Systems are of three main types • Closed System: No mass enters or leaves but heat may be exchanged with the environment • Open System: Mass and energy may enter or leave • Isolated System: Neither mass nor energy may enter or leave L40-s7,8

Physics 114 – Lecture 40 • Calorimetry problems are solved by applying the principle of conservation of energy – heat energy to be specific → • heat lost by one body or bodies = heat gained by the other body or bodies • which is the same statement as • energy lost by one body or bodies = energyt gained by the other body or bodies • Heat lost by one body = m c (Ti – Tf) • Heat gained by one body = m c (Tf – Ti) • Study examples 14.4, 14.5 and 14.6 L40-s8,8

Physics 114 – Lecture 40