First Law of Thermodynamics and Heat Transfer, Exams of Physics

Download First Law of Thermodynamics and Heat Transfer and more Exams Physics in PDF only on Docsity! Chapter 20 – First Law of Thermodynamics This chapter deals with the concept of heat and the First Law of Thermodynamics. Heat is the transfer of energy between a system and its environment because of a temperature difference between them. Internal energy is the energy associated with motions of the atoms and molecules of a system and with the potential energy related to the bonding between atoms or molecules. The molecular motions can be in the form of translation, rotation, and vibration. If heat is transferred from one system to another, then the internal energy of one system will increase and the other will decrease. Units of heat A calorie is the energy required to raise the temperature of 1 g of water from 14.5oC to 15.5oC. 1 cal = 4.186 J One food calorie = 1 Calorie = 1000 calories. (Capital C for food calorie). A Btu is the energy required to raise the temperature of 1 lb of water from 63oF to 64oF. Example: If you consume 2000 Calories of food in a day, then the energy you are consuming is 2000 Calories x 1000 cal/Cal x 4.186 J/cal = 8.37 x 10 6 J Heat Capacity and Specific heat The heat capacity, C, of a substance is the amount of energy, Q, required to raise the temperature by 1 degree. Thus, Q = CT In the SI system, the units of C are J/oC. C is also often expressed in cal/oC. Specific heat is defined as Tm Q c   Units are J/kg·oC or cal/g·oC. Whereas the heat capacity, C, depends on the amount of the material, the specific heat, c, is an intrinsic property of the material and is independent of the amount. By definition of the calorie, for water c = 1 cal/g·oC = 4186 J/kg·oC. Most substances have a smaller specific heat than that of water. From the above equation, the heat required to raise the temperature of something by T is TmcQ  Example: How much heat is required to raise the temperature of 100 g of water from 20oC to 50oC? JxCCkgJkgQ oo 3106.12)30)(/4186)(1.0(  Example: A 500-g block of aluminum initially at 200oC is placed in a bucket containing 1000 g of water initially at 20oC. What would be the final equilibrium temperature of the block and water if heat absorbed by the bucket and heat exchanged with atmosphere could be neglected? The specific heat of aluminum is 0.215 cal/g·oC. C CgcalgCgcalg CCgcalgCCgcalg cmcm CcmCcm T CcmCcmTcmcm CTcmTCcm waterbygainedQAlbylostQ o oo oooo AlAlww o ww o AlAl f o ww o AlAlfAlAlww o fwwf o AlAl 5.37 )/215.0)(500()/1)(1000( )20)(/1)(1000()200)(/215.0)(500( 20200 20200)( )20()200( )()(           Latent heat Latent heat refers to the energy required to cause the phase change of a substance without changing the temperature. Examples of phase changes would be melting, vaporization, sublimation, and change in crystal structure of a solid. Latent heat of fusion is the energy per unit mass to melt a solid. Latent heat of vaporization is the energy per unit mass required to boil a liquid. m Q L  [cal/g, or J/kg] Heat transfer by conduction Consider a slab of material with one face hot and one face cold. The rate of energy transfer from the hot face to the cold face is given by L TT kA x T kA t Q P ch )(        The rate of energy transfer, P, is given in cal/s or J/s (watts). L is the thickness x. The above equation assumes that the faces are at uniform temperature and that there is no heat flow into or out of the sides. The quantity k is the thermal conductivity and is a measure of how easily heat can flow through a material. For example, k for copper would be much larger than for Styrofoam. Example: A room has a wall made of concrete with thickness 15 cm and areal dimensions 2.5 m x 3 m. The inside and outside temperatures are 25oC and 5oC. What is the rate of heat loss through the wall? k(concrete) = 1.3 J/s·m·oC. W m CC mxmCmsJP oo o 1300 15.0 )525( )35.2)(/3.1(    Heat flow through stacked materials For layered materials, the rate of heat flow would be given by    i ii ch kL TT AP / )( where Li and ki and the thickness and thermal conductivity of each layer and Th and Tc are the temperatures of the outside surfaces of the stack of materials. The quantity L/k for a material is called the “R” value. It is a measure of the resistance of a material to the flow of heat. Units used for commercial materials for R are ft2·oF·h/Btu. R depends on the material and its thickness. A good insulating material would have a high R value. For a given thickness, fiberglass would have a much higher R value than wood. Heat transfer by radiation All objects emit heat in the form of electromagnetic radiation. The rate depends on the temperature of the object and the surface area. It also depends on the nature of the surface. The rate of energy transfer (watts) is given by Stefan’s law: 4TeAP   = 5.67 x 10-8 W/m2·K4 is a universal constant A = area of object T is the absolute (Kelvin) temperature. e = emissivity. This depends on the nature of the surface and how well it radiates. e is unitless and e  1. An object will also absorb energy by radiation from its surroundings. If the surrounding temperature is T0, then the net energy radiated will be given by )( 40 4 TTeAP  An object that is a good absorber is also a good radiator. A perfect radiator (and adsorber) is called a ‘blackbody’ and has emissivity e = 1. The electromagnetic radiation of an object occurs over a range of wavelengths. The average wavelength decreases with increasing temperature. Much of the radiation of a hot object may be in the infrared region. Heat transfer by convection An object can also lose or gain heat by the process of convection. This occurs because of the movement of a liquid or gas in the vicinity of the object. For example, air can absorb heat by direct contact with a radiator. This hot air will be expanded and will be buoyed, thus rising and transferring heat elsewhere. You could also have forced convection by blowing air onto a hot or cold object