1 AAST/AEDT AP PHYSICS B: HEAT If we contact two objects with the different temperatures, the hotter one starts to cool and the colder one starts to increase its temperature. The effect can be easily explained. The molecules of the first object have the greater RMS (root mean square) velocity than the molecules of the second object. As the result when molecules of the first object collide with the molecules of the second one, their RMS speed decreases (temperature drops) and the RMS speed of the molecules of the second body increases (temperature rises). The amount of energy transferred during that phenomenon is defined as HEAT. Heat appeared historically, because scientists in the medieval times thought that the heat and the energy are different physical quantities. Today we can study physics and even not use that word, because there is no difference between heat and energy. Off course, because HEAT and Energy are the same, the unit of HEAT is Joule. Sometimes you can find another unit of heat - calorie - it is the amount of energy that is necessary to increase the temperature of one gram of water by 1 degree. That unit had appeared historically. You have to know that: 1 Calorie (Cal)= 4.2 J, or 1 J = 0.24 Cal. The amount of energy that is necessary to increase the temperature of the unit of mass by the unit of temperature is defined as SPECIFIC HEAT. If we use the international system of units - SPECIFIC HEAT is the amount of energy that is necessary to increase the temperature of 1 kg of matter by 1oK. In physics letter C defines Specific heat. That definition means that we can create tables Mass t increase Heat 1 kg - 1 - C m 1 Cm m t Cm t Thus, the formula to estimate the heat necessary to increase the temperature of the mass m by t degrees is Q = Cm t Letter Q defines the amount of the transferred heat. The numerous scientists have measured the data for the specific heat and today we can find it at the physics reference books. Several examples you can see in the tables below
2 Matter Specific heat - Joules/(kg x Kelvin) Aluminum 903 Brass 376 copper 385 glass 664 gold 129 ice 2060 steam 2020 water 4180 Heat transfer is extremely important in technology and we need to know how to estimate the temperature if we are mixing several matters with different temperatures. The simplest example is your shower. You mix hot and cold water. The general procedure of solving of that type of problems is 1. Clarify which objects gain temperature and which lost. 2. Write an expression for the heat gained or lost for every object. 3. Write the heat balance equation - the sum of the gained heat has to be equal to the sum of the lost. 4. Solve the obtained equation respectively to the unknown variable. Example. An iron calorimeter, mass m i =0.1 kg, contains water mass mw=0.5 kg at t 1 = 15oC. The lead block, mass m L = 0.05 kg is dropped into the calorimeter, raising the temperature of the water to = 17 C. Estimate the leads initial temperature. The specific heat of lead is 125.7 J/(kg. K), of iron 460 J/(kg. K), of water 4200 J/(kg. K). m i =0.1 kg In the problem the water and the iron calorimeter increase m w =0.5 kg temperature and the lead lost. So, m L = 0.05 kg Q i =C i m i ( - t 1 )- the amount of heat gained by the calorimeter t 1 = 15 C Q w =C w m w ( - t 1 ) - the amount of heat gained by the water = 17 C Q L =C L m L (t L - ) -the amount of heat lost by the lead C L = 125.7 J/(kg. K) Q i + Q w = Q L - balance equation, or Ci = 460 J/(kg. K) C i m i ( - t 1 ) + C w m w ( - t1) = C L m L (t l - ) Cw = 4200 J/(kg. K) This equation contains only one unknown variable t l -the t l =? initial temperature of the lead. We have to express t l from this equation t 1 = + C i m i ( - t 1 ) + C w m w ( - t 1 ) C L m L Finally we can substitute data t 1 = 17 o + 125.7 J kg K 0.1kg (17 o - 15 o ) + 4200 J kg K 0.5kg (17 o - 15 o ) 199 C 460 J kg K 0.05kg Problem: When a 290 -g piece of iron at 180 C is placed in a 100-g aluminum calorimeter cup, containing 250-g of glycerin at 10 C, the final temperature is observed to be 38 C. What is the specific heat of glycerin?
3 Structure of the matter 1. Gases: Molecules are located far one from another. Their motion is infinite and random. They rarely collide one with another. The collisions are elastic. 2. Solids: Molecules are located at fixed positions and create a regular pattern named crystal lattice. This lattice is different for different matters. Molecules strongly interact with one another. They are vibrating near the fixed position. The amplitude of the vibration is proportional to the temperature. 3. Liquids. Liquid is an intermediate state between solids and gases and the molecular behavior has the features both of the solids and gases. Certain time (the time of the settled life) molecule oscillates near the arbitrary equilibrium position. Then it jumps and starts to vibrate near the another position. The oscillation is the solid type of the molecular motion. These jumps are random. This is the gas type of the molecular motion. If an external force is applied on the liquid, then all jumps are occurred in the same direction and the liquid starts to flow. Change of State Melting:
4 Let us run an experiment: We place a thermometer into the test-tube filled with a solid naphthalene and start to heat the tube. The obtained temperature- time graph is presented on the diagram below. Let us explain this graph. AB- heating of the solid state. The internal structure of the solids is the crystal lattice. When the temperature increases the intensity of the molecular vibrations rises. BC - Melting. The coming energy is set to brake the bonds between the molecules and to destroy the crystal lattice structure. That is why the temperature remains constant during the melting. CD - Heating of the liquid. At that time the intensity of the molecular jumps increases. t 1 - the heating is turned off. DE - Cooling of the liquid EF - Crystallization. Molecules start to catch one another and to create a crystal structure. Energy is emitted during this process. That energy compensates the energy losses to the environment. So, the temperature remains permanent. FG- cooling of the solid. The amount of energy needed to melt the unit mass of the substance at the melting temperature is defined as the heat of fusion. (Latent heat of fusion). The heat of fusion is different for the different matters. You can find the heat of fusion tables in the physics reference books. Example: For ice the heat of fusion is 3.34.10 5 J/kg. The heat of fusion is defined with the letter H f. By the definition for melting 1 kg mass we need the heat that is equal to H f. 1 kg H f If we have mass m, then we will need m times more heat m mh f So, we have Q=mH f
5 The formula allows us to compute the amount of energy that is necessary to melt m kg of the matter, or that is emitted during the crystallization of the same amount of the matter. Example: How much energy we need to obtain a water at t w =40 C from the 5 kg of ice at t i = - 20 C. Solution: First we need to heat the ice from - 20 C to t m = 0 C (melting temperature) Q 1 = C i m( t m - t i ) =2100 J/(kg.K).5 kg.(0 -(-20 ))= 2.1 10 5 J. Then we need to melt the ice. Required energy is: Q 2 = H f m = 3.34 10 5 J/kg 0.5 kg = 16.7.10 5 J The last is to compute the energy needed to heat the water appeared after the ice melts from O C to 40 C. Q 3 = C w m( t w - t m ) =4180 J/(kg.K) 0.5 kg.(40-0 )= 8.36 10 5 J. The total energy needed is Q= Q 1 + Q 2 + Q 3 =2.1 10 5 J + 16.7 10 5 J + 8.36 10 5 J = 27.16 10 5 J VAPORIZATION It is the process of the liquids transformation into the vapor. The opposite phenomenon is named the condensation. The molecules never have the same energies. Some of them have the energy that is above the mean energy. Some of them have the lower energy. If the molecule with a large energy during its random jumps will have a direction towards the liquids surface, it can overcame the attraction of the other molecules and leave the liquid. As the result the mean energy of the remaining molecules starts to decrease and the liquids temperature starts to drop. The more insensitive is the evaporation the greater the cooling. Example: If you wet your hand with an alcohol or with the ether you feel cold. If the evaporation rate is low we often can not observe the cooling effect, because when the liquid temperature drops, the heat from the environment starts to flow to the liquid. That heat compensates the energy losses and the evaporation continues. That heat is called the heat of vaporization. Important difference between the vaporization and the melting is that the vaporization occurs at any temperature. Melting occurs only at one temperature for the particular substance. The amount of energy needed to vaporize the unit mass of the liquid is named the heat of vaporization and labeled H v. If we need to vaporize the mass of m, the necessary heat is Q=mH v Example: How much energy we need to obtain a water steam at t w =100 C from the 5 kg of ice at t i = - 20 C. Solution: First we need to heat the ice from - 20 C to t m = 0 C (melting temperature) Q 1 = C i m( t m - t i ) =2100 J/(kg.K) 0.5 kg.(0 -(-20 ))= 2.1 10 5 J. Then we need to melt the ice. Required energy is:
6 Q 2 = H f m = 3.34 10 5 J/kg 0.5 kg = 16.7 10 5 J The next step is to compute the energy needed to heat the water obtained after the ice had melted from O C to 100 C. Q 3 = C w m( t w - t m ) =4180 J/(kg.K) 0.5 kg.(100-0 )= 20.9 10 5 J. The final step is to calculate the energy needed to convert water at 100 C to the steam at the same temperature. Q 4 = H v m = 2.26 10 6 J/(kg) 0.5 kg= 113 10 5 J. The total energy needed is Q = Q 1 + Q 2 + Q 3 + Q 4 = 2.1 10 5 J + 16.7 10 5 J + 20.9 10 5 J + 113 10 5 J = 152.7 10 5 J Problem: If you mix 100 g of a steam at 100 C with a 200-g of a cold water at 5 C in a container. What final temperature will you observe? What amount of water will be in a container? No heat losses to the environment. Heat of Combustion By definition it is the amount of heat released by the unit of fuel's mass as it burns. It is labeled as H c. If the amount of the burned mass is m than the total heat released is Q= H c m
7 Heat Transfer There are three major methods of heat transfer. They are Conduction, Convection and Radiation Conduction Let us assume that we have a rod with a length of L and the temperature difference at both ends of T and we are able to keep it constant. We define a temperature gradient as the temperature change per unit of length grad T = T/L, or in calculus grad T = dt/dx It can be theoretically and experimentally proved, that the amount of heat that is flowing from the hot end of the rod to the cold one is directly proportional to the Area of the rod s cross-section, to the temperature gradient and to the time of flow. Q ka T L t k is the coefficient that depends on the matter and is named the thermal conductivity. The values for the coefficient you can find in physics reference books. The mechanism for the thermal conductivity is that the hot molecules collide with the cold ones and transform a part of their energy and that is going on like a domino effect. The simplest example of the heating by conduction is the heating of the teaspoon. CONVECTION If you need to heat a room you have to install heater near the floor. At the same time it better to install air conditioner near the ceiling. The reason is that cooling and heating occurs, because of convection. When you heat the air it increases its volume and as the result looses its density. That air experiences a buoyancy force and rises up. Near the ceiling it cools and eventually goes down. The air in the room circulates and makes the room temperature more or less uniform. Radiation
8 In this case the energy is transmitted through the electromagnetic waves. Each object emits electromagnetic waves. Energy radiated by the object can be computed by the formula E= e AT 4 Where e - is the emissivity -coefficient that depends on the object ability to radiate. It varies from 0 (perfect reflector -mirror), to 1 (black body) - is the Stephan Boltzman constant and A is the area of the radiated surface. Example: energy from the sun is coming to the Earth through radiation. Home assignment: Cutnell: Chapter 12, Sections: 12-6 12-8 Conceptual Questions: #11, 13, 16, 17 Problems: Page376 # 41, 45, 48, 52, 57, 63, 65 Chapter 13: page 397 Conceptual questions #2, 4, 9, 11, 15, 19 Problems, Page 398 #3, 5, 9, 10, 13, 27
9 #52-24.958863 6581.89978 Beiser
5 5,1 kj 0.34% 10 80.21 degrees 17 1.36 sec 21 0.35 kg 24 6.51 kw 27.33km/s 37 303 degrees 56 0 degrees 58 0.17 kg 10
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