Chapter 11: Energy in Thermal Process

1. Example homework problems: 11,17,27,38,65 Chapter 11: Energy in Thermal Process • Terminology • Internal energy U is the energy associated with microscopic • components of a system – the atoms and molecules of the system. • It includes kinetic and potential energy associated with the random • translational, and vibrational motion of the particles that make up • the system, and any potential energy bonding the particles together. Heat and Internal Energy • Heat Qis the transfer of energy between a system and its • environment due to a temperature difference between them.

2. Units of heat • Calorie (cal) The energy necessary to raise the temperature of 1 g of water from 14.5o to 15.5oC. “Cal” used for energy content in food is kcal. • British thermal unit (Btu) The energy necessary to raise the temperature of 1 lb of water from 63o to 64oF. Heat and Internal Energy • SI unit : joule (J) 1 cal = 4.186 J

3. Example • Example 11.1 : Working off breakfast Breakfast : two bowls of cereal and milk, 3.20x102 Cal Exercise : Do curls with a 25.0-kg barbell to consume all the calories from the breakfast Breakfast calories: Heat and Internal Energy Work done by one lifting or lowering barbell: Number of repetitions n needed: But see the remark on p.354!

4. Specific heat • Definition of specific heat SI unit : joule per kilo-gram degree Celsius J/(kg oC) DT=Tf-Ti Heat and Internal Energy • The energy required to raise the temp. • of 0.500 kg of water by 3.00oC.

5. Specific heat • Example 11.2 : Stressing a strut A steel strut near a ship’s furnace : L0=2.00 m, ms=1.57 kg, A=1.00x10-4 m2, Q=2.50x105 J. (a) Find the change in temp. of the strut. cs=448 J/(kg oC) Heat and Internal Energy (b) Find the change in length of the strut if it is allowed to expand. a=11x10-6oC-1 (c) Find the compressional stress in the strut. Y=2.00x1011 Pa

6. Isolated system • A system whose energy does not leave out of the system is • called isolated system. • The principle of energy conservation for an isolated system • requires that the net result of all the energy transfer is zero. • If one part of the system loses energy, another part has to • gain the energy. • Calorimeter and calorimetry Calorimetry • Imagine a vessel made of good insulating material and containing • cold water of known mass and temperature and the temperature of • the water can be measure. Such a system of the vessel and water • is called calorimeter. If the object is heated to a higher temperature • of known value before it is put into the water in the vessel, the specific • heat of the object can be measured by measuring the change in • temperature of the water when the system (the object, vessel, and • water) reaches thermal equilibrium. This measuring process is called • calorimetry.

7. Calorimeter and calorimetry • When a warm object, put into a calorimeter with cooler water • described in the previous page, becomes cooler while the water • becomes warmer. • Qcold (>0 ) is the heat transferred (energy change) to the cooler • object and Qhot (<0) is the heat transferred (energy change) to the • warmer object. Calorimetry • In general, in an isolated system consisting of n objects : Tf common to all objects in equilibrium.

8. Calorimeter and calorimetry • Example 11.3 : Finding a specific heat A 125-g block of an unknown substance with a temperature of 90.0oC is placed in a Styrofoam cup containing 0.326 kg of water at 20.0oC. The system reaches an equilibrium temperature of 22.4oC. What is the specific heat, cx , of the unknown substance if the heat capacity of the cup is ignored. Calorimetry

9. Calorimeter and calorimetry • Example 11.4 : Calculate an equilibrium temperature Suppose 0.400 kg of water initially at 40.0oC is poured into a 0.300-kg glass beaker having a temperature of 25.0oC. A 0.500-kg block of aluminum at 37.0oC is placed in the water, and the system is insulated. Calculate the final equilibrium temperature of the system. Calorimetry

10. Latent heat and phase change • Phase change (change of molecular structure) - Phases of material : solid, liquid, gas - A change of phase : phase transition - For any given pressure a phase change takes place at a definite temperature, usually accompanied by absorption or emission of heat and a change of volume and density Heated added and phase changes of water Calorimetry

11. Latent heat and phase change • Latent heat - To change structure of material, change in energy required. • When a substance goes under phase transition, all the heat • transferred is used to change the phase. L : latent heat (J/kg) , m : mass (kg) +(-) when energy is absorbed (removed). Calorimetry Heated added and phase changes of water

12. Latent heat and phase change • Latent heat • The latent heat of fusion Lf is used when a phase change occurs • during melting or freezing. • The latent heat of vaporization Lv is used when a phase change • occurs during boiling or condensing. Calorimetry

13. Latent heat and phase change • Water Consider an addition of energy to a 1.00-g cube of ice at -30.0oC in a container held at constant pressure. Suppose this input energy turns ice to steam (water vapor) at 120.0oC. A: Calorimetry B: C: D: E:

14. Latent heat and phase change • Example 11.5 : Boiling liquid helium Liquid helium has a very low boiling point, 4.2 K, as well as a low latent heat of vaporization, 2.09x104 J/kg. If energy is transferred to a container of liquid helium at the boiling point from an immersed electric heater at a rate of 10.0 W, how long does it take to boil away 2.00 kg of the liquid? Calorimetry

15. Latent heat and phase change • Example 11.6 : Ice water 6.00 kg of ice at -5.00oC is added to a cooler holding 30 liters of water at 20.0oC. What temperature of the water when it comes to equilibrium? Q m(kg) c(J/(kgoC)) L (J/kg) Tf Ti Exp. Qice 6.00 2090 0 -5.00 mcDT Qmelt 6.00 3.33x105 0 0 mLf Calorimetry Qice-water 6.00 4190 T 0 mcDT Qwater 30.0 4190 T 20.0 mcDT

16. Latent heat and phase change • Example 11.7 : Ice water A 5.00-kg block of ice at 0oC is added to an insulated container partially filled with 10.0 kg of water at 15.0oC. (a) Find the final temperature (neglect the heat capacity of the container.) The amount of energy needed to completely melt the ice: The maximum energy that can be lost by the initial mass of liquid water without freezing it : Calorimetry Since this is less than half the energy needed to melt all the ice, so the final state of the system is a mixture of water and ice at the freezing point. (b) Find the mass of the ice melted.

17. Mechanism of heat transfer • Thermal conduction • Convection • Radiation • Thermal conduction • One form of heat transfer when there is temperature difference is • thermal conduction. • This process is caused by an exchange of kinetic energy between • microscopic particles such as molecules, atoms, and electrons etc. • In the process, less energetic particles gain energy as they collide • with more energetic particles. Energy Transfer The rate of energy transfer P=Q/Dt is proportional to cross-sectional area A, the difference in temperature DT=Th-Tc>0, and inversely proportional to the thickness of the slab.

18. Thermal conduction • Consider a substance is in the shape of a long, uniform rod of length • L and assume that the rod is insulated so that thermal energy does not • escape from the surface. k : proportionality constant thermal conductivity [J/(s m oC)] Energy Transfer Th : higher temperature (oC) Tc : lower temperature (oC) A : area (m2) L : length (m) Q : heat transferred (J) Dt : time interval (s)

19. Thermal conduction • Thermal conductivities • Example 11.9 : Energy transfer through a concrete • wall Find the energy transferred in 1.00 h by conduction through a concrete wall 2.0 m high, 3.65 m long, and 0.20 m thick if one side of the wall is held at 20oC and the other at 5oC. Energy Transfer

20. Thermal conduction • Example : Two rods cases k1,L,,A1 k2,L2 k1,L1 Tm k2,L,,A2 Energy Transfer

21. Thermal conduction • Thermal conduction through multiple layers L2 L3 L1 k2 k3 k1 Energy Transfer Th2 Th3 Th3 Th1 Tc1 Tc2 Tc2 Th

22. Thermal conduction • Example 11.11 : Staying warm in the arctic An arctic explore builds a wooden shelter out of wooden planks that are 1.0 cm thick. To improve the insulation, he covers the shelter with a layer of ice 3.2 cm thick. (a) Compute the R factors for the wooden planks and the ice. (b) Find the rate of heat loss. Energy Transfer (c) Find the temperature in between the ice and the wood.

23. Convection • The transfer of energy by the movement of a substance is called • convection. Energy Transfer

24. Radiation • All objects radiate energy because of microscopic movements • (accelerations) of charges, which increase with temperature. Heat current in radiation e : emissivity that depends on nature of surface T in K A :area Energy Transfer s=5.670400(40) x 10-8 W/(m2 K4) :Stefan-Boltzman const. • If an object is at temperature T and its surroundings are at • temperature T0, the net flow of heat radiation between the object and its surroundings is Heat transfer by radiation

25. Radiation • Example 11.12 : Polar bear club A member of the Polar Bear Club, dressed only in bathing trunks of negligible size, prepares to plunge into the Baltic Sea from the in St. Petersburg, Russia. The air is calm, with a temperature of 5oC. If the swimmer’s surface body temperature is 25oC, compute the net rate of energy loss from his skin due to radiation. How much energy is lost in 10.0 min? Assume his emissivity is 0.900 and his surface area is 1.50 m2. Energy Transfer