Lecture 25 Thermodynamics, Heat and Temp (cont.) Heat and temperature Gases & Kinetic theory http://candidchatter.files.wordpress.com/2009/02/hell.jpg
Specific Heat Specific Heat: heat capacity per unit mass Q C T cm T Specific Heat of Water: 1 cal ggc o 4190 J kggk
Specific Heat After a grueling work out, you drink a liter of cold water (0C). How many calories does it take for the body to raise the water up to a body temperature of 36 C? Specific Heat of Water: 1 cal ggc o 4190 J kggk 1 liter 1000g of H 2 O 1000g*1 cal gc *36C 36,000cal 36 Calories 1 pat of butter is 36 Calories 1 slice of bread (0.4 oz no crust) ~ 32 Calories
Quiz: Two blocks of equal mass but different material are placed in thermal contact. One is at 100 o C and the other at 0 o C. What temperature do they reach at thermal equilibrium? A. 100 o C B. 50 o C C. 0 o C D. Can t tell
Phase Transitions As you add heat to water, the temperature increases for a while, then it remains constant, despite the addition of heat! Q=mL Latent Heat, L, (J/kg) is the heat which must be added to (or removed from) 1 kg of a material to change from one phase to another phase. During the conversion the temperature remains constant.
Phase Transitions Latent Heat of Fusion, L F, : Solid to Liquid Latent Heat of Vaporization, L V, : Liquid to Gas Latent Heat of Sublimation, L s, : Solid to Gas
Phase Transitions
System complete converted to steam at 138 C. To make steam, you add Example 5.60 x10 5 J of heat to 0.220 kg of water at an initial temperature of 50.0 o C. Find the final temperature of the steam. Calculate heat needed to heat water to 100C: Q 1 mc water T 0.220kg *4186 J kg * C o *50.0C 4.60x104 J Calculate heat needed to convert liquid to steam: Q 2 ml v 0.220kg * 22.6x10 5 J kg 4.97x105 J Determine Heat that is still available to add to system Q 3 5.60x10 5 J Q 1 Q 2 5.60x10 5 J 4.60x104 4.97x10 5 J 17,000J Use Q 3 to find increase in temperature above 100C: Q 3 17,000J mc steam T T Q 3 mc steam 17,000J J 0.220kg *2010 kg *C o o 38C
Chapter 15: Gases & Kinetic Theory
Avogadro s Number One mole = number of atoms in a 12g sample of carbon 12. How many atoms/molecules are in a mole? Avogadro s Number: N A 6.02x10 23 1 mol Number of moles, n, contained in a sample: n N N A N=number of molecules in a sample
Atomic Mass
Molar Mass mass of molecule in u is numerically the same as the molar mass g/mole. Ex. O 2 molar mass: 31.998 g/mole mass of one molecule: 31.998 u
Ideal Gas Law PV NkT P: pressure V: volume N nn A N: number of molecules k: Boltzman s constant = 1.38 x10-23 J/K T: Temperature k R N A nr Nk
Ideal Gas Law PV nrt P: pressure V: volume N nn A n: number of moles R: gas constant = 8.31 J/mol*K T: Temperature
Example One mole (1.00 mol) of Helium gas is at a room temperature of 300 K and 1.00 atm of pressure. How many Helium atoms are in each cubic millimeter of the gas? PV nrt J V nrt 1.00mol * 8.315 P molgk *300K 1.013x10 5 Pa V 2.46x10 2 m 3 One cubic mm = 10-9 m 3 N V ' V N A 1.0x10 9 m 3 2.46x10 2 m 3 *6.02x1023 atoms 2.45x10 16 atoms
Ideal Gas Law PV nrt For Constant Number of molecules: P 1 V 1 T 1 P 2V 2 T 2 For Constant Number of molecules & Temperature Boyle s Law P 1 V 1 P 2 V 2
Ideal Gas Law For Constant n & T Boyle s Law P 1 V 1 P 2 V 2 http://charles-boyle-law.org.ru/im/im5.jpg
Example In the morning, when the temperature is 288 K, a bicyclist finds that the absolute pressure in the tires is 505 kpa. That afternoon, she finds that the pressure in the tires has increased to 552 kpa. Ignoring expansion of the tires, find the afternoon temperature. P 1 V 1 T 1 nr const. P 2V 2 T 2 P 1 T 1 P 2 T 2 T 2 P 2 T 1 P 1 552kPa *288K 505kPa 315K
Kinetic Theory of Gases Kinetic energy of a molecule KE 3 2 k B T Kinetic energy of a molecule KE mv 2 2 Average velocity v 3k B m T
Maxwell s Speed Distribution v 3k B m T
Molecular Speed: v rms 3kT m Escape Speed: v e 2GM E R E 11,200m / s
If the temperature of an ideal gas is doubled and the pressure is held constant, the average speed of the molecules A. remains unchanged B. is 4 times the original speed 2 C. is times the original speed D. is 2 times the original speed