Thermal Properties of Matter

1. How Hot Things Behave Thermal Properties of Matter

2. Objectives • Define specific heat capacity & thermal capacity. • Solve problems involving specific heat capacities and thermal capacities. • Explain the physical differences between the solid, liquid and gaseous phases in terms of molecular structure and particle motion. • Describe and explain the process of phase changes in terms of molecular behavior. • Explain in terms of molecular behavior why temperature does not change during a phase change. • Distinguish between evaporation and boiling.

3. Define specific heat capacity & thermal capacity Specific Heat Capacity: We know that temperature is a measure of the average KE per particle in a substance. However not all substances will behave in the same way when heated. The specific heat capacity of a material is the amount of energy (in Joules) needed to increase the temperature of one kilogram of mass of the material by one Kelvin. ∆Q = mc ∆T ∆Q = Amount of heat energy (J) ∆T = Change in temp (K) m = mass (kg) c = specific heat capacity(Jkg-1K-1) Units: J kg-1 K-1

4. Define specific heat capacity & thermal capacity. How to read this table: The thermal energy required to increase the temperature of 1 kg of Iron by 1 K is 470 J

5. Solve problems involving specific heat capacities and thermal capacities. Q. Explain the following in terms of particles… (you may wish to remind yourself of the definition of absolute temperature)… • To raise the temperature of 1 kg of lead by one Kelvin requires 130J, whereas for 1 kg of copper 390J are needed. • To raise the temperature of one mole of lead or one mole of copper by one Kelvin requires about the same amount of energy.

6. Heat Capacity Whereas ‘specific’ heat capacity is ‘per kg’ of a material, heat capacity (or thermal capacity) is the energy required to raise a certain objects temperature by one Kelvin, irrespective of its mass. Heat capacity = mass x specific heat capacity C = mc m = mass (kg) C = heat capacity (JK-1) c = specific heat capacity(Jkg-1K-1) ∆T = Change in temp (K) ∆Q = Amount of heat energy (J) ∆Q = C∆T

7. Solve problems involving specific heat capacities and thermal capacities. E.g. What is the heat capacity (thermal capacity) of 5kg of sand? How much heat is required to raise the temperature of the sand by 7.5°C (csand = 835 Jkg-1K-1)? A. C = mc = 5 x 835 = 4175 JK-1 ∆Q = C∆T = 4175 x 7.5 = 31312 J

8. Solve problems involving specific heat capacities and thermal capacities. Note: A glass beaker containing water has a heat capacity that is equal to that of the water plus that of the glass. So for such an object… Alternatively, if we assume the glass and water increase by the same change in temperature, we can say the total energy given to the glass plus water is given by… C = m1c1 + m2c2 Where 1 and 2 represent the two materials making up an object. ∆Q = m1c1∆T + m2c2∆T

9. Experimental Determination of Specific Heat Capacity Note: To minimise error… • Low mass thermometer and heater • Insulation • Wait for max temp after turning off • Stir liquids Also… Consider s.h.c. of calorimeter (beaker / container)

10. Objectives • Define specific latent heat. • Solve problems involving specific latent heats. • Define pressure. • State that temperature is a measure of the average random kinetic energy of the molecules of an ideal gas. • Explain the macroscopic behavior of an ideal gas in terms of a molecular model.

11. Specific Latent Heat If a substance is heated to its melting or boiling point, any further heatsupplied to the substance will be used to do workon the molecules, overcoming the forces that hold them together and causing a change of state. This all occurs at a constant temperature. The heat supplied to cause this change of state is called latent heat. The specific latent heat of a material is the amount of energy needed to change one kilogram of mass of the material from one state of matter to another, without change in temperature.

12. Specific Latent Heat The thermal energy required to melt a unit mass of material at its melting point is called specific latent heat of fusion , Lfand the thermal energy required to vaporize a unit mass at its boiling point is called the specific latent heat of vaporization , Lv. Thus to melt or vaporize a quantity of mass m, we require a quantity of thermal energy Q = mLf andQ = m Lv Units: J kg-1

13. Solve problems involving specific heat capacities and thermal capacities. Q.Explain why supply of latent heat causes a change in potential energy but not kinetic energy. NOTE: We are interested in… i. Specific latent heat of vaporisation (Lv) - from liquid to gas ii. Specific latent heat of fusion (Lf) - from solid to liquid A. Kelvin temperature is proportional to the KE of the molecules. Thus if the temperature hasn’t increased then the KE has also not increased. Work has been done against forces, changing the position of the molecules. Thus the PE of the molecules must change.

14. Heat supplied = mass of substance x specific latent heat ∆Q = mLv ∆Q = Amount of heat energy (J) L = Specific latent heat (Jkg-1) m = mass (kg) So… and… E.g. 0.5kg of ice at 0°C is heated… i. … until it has all melted to water at 0°C ii. … until it reaches it’s boiling point at 100°C iii. … until it has vaporised to steam at 100°C Calculate the heat supplied at each stage and in total. ∆Q = mLf

15. Continued… Ai. ΔQ = mLf = 0.5 x 334 x 103 = 1.67 x 105 J ii. ΔQ = mcΔT = 0.5 x 4200 x 100 = 2.10 x 105 J iii.ΔQ = mLv = 0.5 x 2258 x 103 = 1.13 x 106 J Total = 1.51 x 106 J ( = 1.51 MJ)

16. Pressure • Force per unit area • Equation: • Units: N m-2 = Pascal (Pa) • Gas pressure caused by collisions with container • Other units for pressure • 1 bar - 100,000 Pa • 1 millibar - 100 Pa • 1 atmosphere - 101,325 Pa • 1 mm Hg - 133 Pa • 1 inch Hg - 3,386 Pa • Common Pressures: • Atmospheric pressure @ sea level = 101.325 kPa • 10 Pa - the pressure below 1 mm of water • 1 kPa - approximately the pressure exerted by a 10 g of mass on a 1 cm2 area • 10 kPa - the pressure below 1 m of water, or the drop in air pressure when moving from sea level to 1000 m elevation • 10 MPa - nozzle pressure in a "high pressure" washer • 10 GPa - pressure enough to form diamonds

17. Explain the physical differences between the solid, liquid and gaseous phases in terms of molecular structure and particle motion. Particles in a solid are locked into the crystal structure and do not have enough KE to escape. In a liquid, they overcome the chemical bonds but cannot separate completely for the other atoms. The particles in a gas have enough KE to escape from each other.

18. Describe and explain the process of phase changes in terms of molecular behavior. Once at the melting/boiling point, any additional thermal energy supplied does not increase the temperature. Rather, this thermal energy is used to overcome the forces between molecules and separate them (work).

19. Explain in terms of molecular behavior why temperature does not change during a phase change. • During a phase change, heat (energy) is still being added to/removed from the system, but instead of being used to increase/decrease the kinetic energy of the particles (temperature), it’s being used to do work on the particles to separate them. • Intermolecular forces dominate in solids, are weak in liquids and practically non-existent in gasses.

20. Distinguish between evaporation and boiling • Boiling occurs at a fixed temperature while evaporation occurs at any temperature • During boiling, molecules can escape from anywhere in the liquid. Evaporation occurs only at the surface. More surface area = faster evaporation • Specific heat of vaporization affects evaporation rate