Physics 101 Lecture 5 (Heat) PDF

Faculty of Women Physics Department Physics 101 (heat) Lecture 5 Prepared By: Dr. Doaa Abdallah Reference: Peter J. Nolan - Fundamentals of College Physics, Vol. 1, 5th Updated Edition (2005, Pearson Custom Publishing)...

Faculty of Women Physics Department Physics 101 (heat) Lecture 5 Prepared By: Dr. Doaa Abdallah Reference: Peter J. Nolan - Fundamentals of College Physics, Vol. 1, 5th Updated Edition (2005, Pearson Custom Publishing)  Gay-Lussac’s Law Consider a gas contained in a tank, as shown in figure (1). The tank is made of steel and there is a negligible change in the volume of the tank, and hence the gas, as it is heated. A pressure gauge attached directly to the tank, is calibrated to read the absolute pressure of the gas in the tank. A thermometer reads the temperature of the gas in degrees Celsius. The tank is heated, thereby increasing the temperature and the pressure of the gas, which are then recorded. If we plot the pressure of the gas versus the temperature, we obtain the graph of figure (2). Figure (1) Changing the Figure (2) A plot of pressure versus pressure of a gas. temperature for a gas. ▪ The equation of the resulting straight line is: p − p0 = m’(t − t0) where p is the pressure of the gas at the temperature t, p0 is the pressure at the temperature t0, and m’ is the slope of the line. Because t0 = 0 0C, this simplifies to: p − p0 = m’t or p = m’t + p0 (1) ▪ It is found experimentally that the slope is : m’ = p0β where p0 is the absolute pressure of the gas and β is the coefficient of volume expansion for a gas. Therefore equation (1) becomes: p = p0βt + p0  p = p0(βt + 1) (2) ▪ Thus, the pressure of the gas is a linear function of the temperature, as in the case of Charles’ law. Since β = 1/273 0C this can be written as: (3) ▪ But the absolute or Kelvin scale has already been defined as: T K = t 0C + 273 ▪ Therefore, equation (3) becomes: (4) which shows that the absolute pressure of a gas at constant volume is directly proportional to the absolute temperature of the gas, a result known as Gay-Lussac’s law, in honor of the French chemist Joseph Gay-Lussac (1778-1850). ▪ For a gas in different states at two different temperatures, we have: And (5) Equation (5) is another form of Gay-Lussac’s law.  Boyle’s Law Consider a gas contained in a cylinder at a constant temperature, as shown in figure (3). By pushing the piston down into the cylinder, we increase the pressure of the gas and decrease the volume of the gas. Figure (3) The change in pressure and volume of a gas at constant temperature. If the pressure is increased in small increments, the gas remains in thermal equilibrium with the temperature reservoir, and the temperature of the gas remains a constant.  We measure the volume of the gas for each increase in pressure and then plot the pressure of the gas as a function of the reciprocal of the volume of the gas. The result is shown in figure (4). Figure (4) Plot of the pressure p versus the reciprocal of the volume 1/V for a gas. ▪ Notice that the pressure is inversely proportional to the volume of the gas at constant temperature. We can write this as: That is, the product of the pressure and volume of a gas at constant temperature is equal to a constant, a result known as Boyle’s law, in honor of the British physicist and chemist Robert Boyle (1627-1691). ▪ For a gas in two different equilibrium states at the same temperature, we write this as : p1V1 = constant and p2V2 = constant Therefore, p1V1 = p2V2 T = constant (7) another form of Boyle’s law.  The Ideal Gas Law ▪ The three gas laws, can be combined into one equation, namely: (8) Equation (8) is a special case of a relation known as the ideal gas law. Hence, we see that the three previous laws, which were developed experimentally, are special cases of this ideal gas law, when either the pressure, volume, or temperature is held constant. The ideal gas law is a more general equation in that none of the variables must be held constant. Equation (8) expresses the relation between the pressure, volume, and temperature of the gas at one time, with the pressure, volume, and temperature at any other time. For this equality to hold for any time, it is necessary that:  (9) This constant must depend on the quantity or mass of the gas. A convenient unit to describe the amount of the gas is the mole. One mole of any gas is that amount of the gas that has a mass in grams equal to the atomic or molecular mass (M) of the gas. The mole is a convenient quantity to express the mass of a gas because one mole of any gas at a temperature of 0 0C and a pressure of 1 atmosphere, has a volume of 22.4 liters. Also Avogadro’s law states that every mole of a gas contains the same number of molecules. This number is called Avogadro’s number NA and is equal to 6.022 × 1023 molecules/mole. ▪ The mass of any gas will now be represented in terms of the number of moles, n. We can write the constant in equation (9) as n times a new constant, which shall be called R, that is: (10) ▪ To determine this constant R let us evaluate it for 1 mole of gas at a pressure of 1 atm and a temperature of 0 0C, or 273 K, and a volume of 22.4 L. That is: ▪ Converted to SI units, this constant is: ▪ We call the constant R the universal gas constant, and it is the same for all gases. We can now write equation (9) as pV = nRT (11) Equation (11) is called the ideal gas equation. Remember that the temperature T must always be expressed in Kelvin units. ❖ Example (1) The pressure of an ideal gas is kept constant while 3.00 m3 of the gas, at an initial temperature of 50.0 0C, is expanded to 6.00 m3. What is the final temperature of the gas? Solution The temperature must be expressed in Kelvin units. Hence the initial temperature becomes T1 = t 0C + 273 = 50.0 + 273 = 323 K We find the final temperature of the gas by using the ideal gas equation in the form of equation (7), namely: However, since the pressure is kept constant, p1 = p2, and cancels out of the equation. Therefore: and the final temperature of the gas becomes: ❖ Example (2) A balloon is filled with helium at a pressure of 2.03 × 105 N/m2, a temperature of 35.0 0C, and occupies a volume of 3.00 m3. The balloon rises in the atmosphere. When it reaches a height where the pressure is 5.07 × 104 N/m2, and the temperature is −20.0 0C, what is its volume? Solution First we convert the two temperatures to absolute temperature units as: T1 = 35.0 0C + 273 = 308 K and T2 = −20.0 0C + 273 = 253 K We use the ideal gas law in the form: Solving for V2 gives, for the final volume: ❖ Example (3) What is the pressure produced by 2.00 moles of a gas at 35.0 0C contained in a volume of 5.00 × 10−3 m3? Solution We convert the temperature of 35.0 0C to Kelvin by: T = 35.0 0C + 273 = 308 K We use the ideal gas law in the form: pV = nRT Solving for p: ❖ Example (4) Compute the number of molecules in a gas contained in a volume of 10.0 cm3 at a pressure of 1.013 × 105 N/m2, and a temperature of 30 K. Solution The number of molecules in a mole of a gas is given by Avogadro’s number NA, and hence the total number of molecules N in the gas is given by: N = nNA Therefore we first need to determine the number of moles of gas that are present. From the ideal gas law: pV = nRT The number of molecules is now found as:  Heat Transfer ▪ An amount of thermal energy Q, given by: Q = mc∆T is absorbed or liberated in a sensible heating process. But how is this thermal energy transferred to, or from, the body so that it can be absorbed, or liberated? To answer that question, we need to discuss the mechanism of thermal energy transfer. Thermal energy can be transferred from one body to another by any or all of the following mechanisms: 1. Convection 2. Conduction 3. Radiation  Convection is the transfer of thermal energy by the actual motion of the medium itself. The medium in motion is usually a gas or a liquid. Convection is the most important heat transfer process for liquids and gases. Conduction is the transfer of thermal energy by molecular action, without any motion of the medium. Conduction can occur in solids, liquids, and gases, but it is usually most important in solids. Radiation is a transfer of thermal energy by electromagnetic waves. It is not necessary to have a medium for the transfer of energy by radiation. For example, energy is radiated from the sun as an electromagnetic wave, and this wave travels through the vacuum of space, until it impinges on the earth, thereby heating the earth.  Convection Consider the large mass m of air at the surface of the earth that is shown in figure (5) The lines labeled T0, T1, T2, and so on are called isotherms and represent the temperature distribution of the air at the time t. Figure (5) Horizontal convection. An isotherm is a line along which the temperature is constant.  Thus, everywhere along the line T0 the air temperature is T0, and everywhere along the line T1 the air temperature is T1, and so forth. Consider a point P on the surface of the earth that is at a temperature T0 at the time t. How can thermal energy be transferred to this point P thereby changing its temperature? That is, how does the thermal energy at that point change with time? If we assume that there is no local infusion of thermal energy into the air at P, such as heating from the sun and the like, then the only way that thermal energy can be transferred to P is by moving the hotter air, presently to the left of point P, to point P itself. That is, if energy can be transferred to the point P by convection, then the air temperature at the point P increases. ▪ The transfer of thermal energy per unit time to the point P is given by ∆Q/∆t. By multiplying and dividing by the distance ∆x, we can write this as: (1) But ▪ the velocity of the air moving toward P. Therefore, equation (1) becomes: (2) ▪ But ∆Q, on the right-hand side of equation (2), can be replaced with: ∆Q = mc∆T Therefore: (3) ▪ Hence, the thermal energy transferred to the point P by convection becomes (4) The term ∆T/∆x is called the temperature gradient, and tells how the temperature changes as we move in the x-direction. ❖ Example (5) If the temperature gradient is 2.00 0C per 100 km and if the specific heat of air is 1009 J/(kg 0C), how much thermal energy per unit mass is convected to the point P in 12.0 hr if the air is moving at a speed of 10.0 km/hr? Solution The heat transferred per unit mass, found from equation (4), is: If the mass m of the air that is in motion is unknown, the density of the fluid can be used to represent the mass. Because the density ρ = m/V, where V is the volume of the air, we can write the mass as: m = ρV (5) Therefore, the thermal energy transferred by convection to the point P becomes: (6) Sometimes it is more convenient to find the thermal energy transferred per unit volume. In this case, we can use equation (6) as: ❖ Example (6) If the temperature gradient is 2.00 0C per 100 km and if the specific heat of air is 1009 J/(kg 0C), find the thermal energy per unit volume transferred by convection to the point P in 12.0 hr Assume that the density of air is ρair = 1.293 kg/m3, if the air is moving at a speed of 10.0 km/hr? Solution The thermal energy transferred per unit volume is found as :  Transfer of thermal energy by convection is also very important in the process called the sea breeze, which is shown in figure 16.3. Water has a higher specific heat than land and for the same radiation from the sun, the temperature of the water does not rise as high as the temperature of the land. Therefore, the land mass becomes hotter than the neighbouring water. The hot air over the land rises and a cool breeze blows off the ocean to replace the rising hot air. Air aloft descends to replace this cooler sea air and the complete cycle is as shown in figure (6). The net result of the process is to replace hot air over the land surface by cool air from the sea. This is one of the reasons why so many people flock to the ocean beaches during the hot summer months. The process reverses at night when the land cools faster Figure (6) The sea breeze. than the water. The air then flows from the land to the sea and is called a land breeze.  This same process of thermal energy transfer takes place on a smaller scale in any room in your home or office. Let us assume there is a radiator situated at one wall of the room, as shown in figure (7). The air in contact with the heater is warmed, and then rises. Cooler air moves in to replace the rising air and a convection cycle is started. Figure (7) Natural convection in a room. The net result of the cycle is to transfer thermal energy from the heater to the rest of the room. All these cases are examples of what is called natural convection.  Conduction Conduction is the transfer of thermal energy by molecular action, without any motion of the medium. Conduction occurs in solids, liquids, and gases, but the effect is most pronounced in solids. If one end of an iron bar is placed in a fire, in a relatively short time, the other end becomes hot. Thermal energy is conducted from the hot end of the bar to the cold end. The atoms or molecules in the hotter part of the body vibrate around their equilibrium position with greater amplitude than normal. This greater vibration causes the molecules to interact with their nearest neighbours, causing them to vibrate more also. These in turn interact with their nearest neighbours passing on this energy as kinetic energy of vibration. The thermal energy is thus passed from molecule to molecule along the entire length of the bar. The net result of these molecular vibrations is a transfer of thermal energy through the solid.  Heat Flow Through a Slab of Material We can determine the amount of thermal energy conducted through a solid with the aid of figure (8). A slab of material of cross-sectional area A and thickness d is subjected to a high temperature Th on the hot side and a colder temperature Tc on the other side. Figure (8) Heat conduction through a slab.  It is found experimentally that the thermal energy conducted through this slab is directly proportional to: (1) the area A of the slab — the larger the area, the more thermal energy transmitted. (2) the time t — the longer the period of time, the more thermal energy transmitted. (3) the temperature difference, Th − Tc, between the faces of the slab. If there is a large temperature difference, a large amount of thermal energy flows. ▪ We can express these observations as the direct proportion: Q ∝ A(Th − Tc)t ▪ The thermal energy transmitted is also found to be inversely proportional to the thickness of the slab, that is: This is very reasonable because the thicker the slab the greater the distance that the thermal energy must pass through. Thus, a thick slab implies a small amount of energy transfer, whereas a thin slab implies a larger amount of energy transfer. ▪ These two proportions can be combined into one as: (7) ▪ To make an equality out of this proportion we must introduce a constant of proportionality. The constant must also depend on the material that the slab is made of, since it is a known fact that different materials transfer different quantities of thermal energy. We will call this constant the coefficient of thermal conductivity, and will denote it by k. Equation (7) becomes: (8) Equation (8) gives the amount of thermal energy transferred by Conduction. Table (1) gives the thermal conductivity k for various materials. If k is large, then a large amount of thermal energy will flow through the slab, and the material is called a good conductor of heat. If k is small then only a small amount of thermal energy will flow through the slab, and the material is called a poor conductor or a good insulator.  Note from table (1) that most metals are good conductors while most nonmetals are good insulators. ❖ Example (7) Heat transfer by conduction. Find the amount of thermal energy that flows per day through a solid oak wall 10.0 cm thick, 3.00 m long, and 2.44 m high, if the temperature of the inside wall is 21.1 0C while the temperature of the outside wall is −6.67 0C. Solution The thermal energy conducted through the wall, found from equation (8), is:  Radiation Radiation is the transfer of thermal energy by electromagnetic waves. The electromagnetic waves in the visible portion of the spectrum are called light waves. These light waves have wavelengths that vary from about 0.38 × 10−6 m for violet light to about 0.72 × 10−6 m for red light. Above visible red light there is an invisible, infrared portion of the electromagnetic spectrum. Most, but not all, of the radiation from a hot body falls in the infrared region of the electromagnetic spectrum. Every thing around you is radiating electromagnetic energy, but the radiation is in the infrared portion of the spectrum, which your eyes are not capable of detecting. Therefore, you are usually not aware of this radiation.  The Stefan-Boltzmann Law Joseph Stefan (1835-1893) found experimentally, and Ludwig Boltzmann (1844-1906) found theoretically, that: every body at an absolute temperature T radiates energy that is proportional to the fourth power of the absolute temperature. ▪ The result, which is called the Stefan-Boltzmann law is given by: Q = eσAT4t (9) where Q is the thermal energy emitted; e is the emissivity of the body, which varies from 0 to 1; σ is a constant, called the Stefan-Boltzmann constant and is given by: A is the area of the emitting body, T is the absolute temperature of the body, and t is the time.  Radiation from a Blackbody The amount of radiation depends on the radiating surface. Polished Surfaces are usually poor radiators, while blackened surfaces are usually good radiators. Good radiators of heat are also good absorbers of radiation, while poor radiators are also poor absorbers. A body that absorbs all the radiation incident upon it is called a blackbody. The name blackbody is really a misnomer, since the sun acts as a blackbody and it is certainly not black. A blackbody is a perfect absorber and a perfect emitter. For a blackbody, the emissivity e in equation (9) is equal to 1. ▪ The amount of heat absorbed or emitted from a blackbody is: Q = σAT 4t ❖ Example (8) Energy radiated from the sun. If the surface temperature of the sun is approximately 5800 K, how much thermal energy is radiated from the sun per unit time? Assume that the sun can be treated as a blackbody, The radius of the sun is about 6.96 × 108 m Solution We can find the energy radiated from the sun per unit time from equation (9). e = 1 for a blackbody Its area is therefore: A = 4πr2 = 4π(6.96 × 108 m)2 = 6.09 × 1018 m2 The heat radiated from the sun is therefore: