Lecture 25 – , HeatHeat and and Temp (cont.)

Gases & Kinetic theory

http://candidchatter.files.wordpress.com/2009/02/hell.jpg Physics 218 Fall 2010 Specific Specific Heat: per unit mass Q  CT  cmT

Specific Heat of Water:

cal J 1  4190 ggC o kggK

Physics 218 Fall 2010 Specific Heat After a grueling 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? cal J Specific Heat of Water: 1  4190 ggC o kggK

1 liter  1000g of H2O cal 1000g*1 *36C  36,000cal  36 Calories gC

1 pat of butter is 36 Calories 1 slice of bread (0.4 oz no crust) ~ 32 Calories

Physics 218 Fall 2010 Quiz: Two blocks of equal mass but different material are placed in thermal contact. One is at 100 oC and the other at 0 oC. What temperature do they reach at thermal equilibrium?

A. 100 oC B. 50 oC C. 0 oC D. Can’t tell

Physics 218 Fall 2010 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. Physics 218 Fall 2010 Phase Transitions

Latent Heat of Fusion, LF, : to

Latent Heat of Vaporization, LV, : Liquid to

Latent Heat of Sublimation, Ls , : Solid to Gas

Physics 218 Fall 2010 Phase Transitions

Physics 218 Fall 2010 System complete converted to steam at 138 C. To make steam, you addExample 5.60 x105 J of heat to 0.220 kg of water at an initial temperature of 50.0 oC. Find the final temperature of the steam.

Calculate heat needed to heat water to 100C:

J Q  mc T  0.220kg *4186 *50.0C  4.60x104 J 1 water kg *C o Calculate heat needed to convert liquid to steam:

J Q  mL  0.220kg * 22.6x105  4.97x105 J 2 v kg Determine Heat that is still available to add to system 5 5 5 Q3  5.60x10 J  Q1  Q2  5.60x10 J  4.60x104  4.97x10 J  17,000J

Use Q3 to find increase in temperature above 100C: Q3  17,000J  mcsteam T Q 17,000J T  3   38C o J mcsteam 0.220kg *2010 o Physics 218 Fall 2010 kg *C Chapter 15: & Kinetic Theory

Physics 218 Fall 2010 Avogadro’s Number One = number of atoms in a 12g sample of carbon 12.

How many atoms/molecules are in a mole? 1 Avogadro’s Number: N  6.02x1023 A mol

Number of moles, n, contained in a sample: N n  N=number of molecules in a sample N A

Physics 218 Fall 2010 Atomic Mass

Physics 218 Fall 2010 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

Physics 218 Fall 2010 PV  NkT

P: V: N  nN A N: number of molecules k: Boltzman’s constant = 1.38 x10-23 J/K R T: Temperature k  nR  Nk N A

Physics 218 Fall 2010 Ideal Gas Law PV  nRT

P: pressure V: volume N  nN A n: number of moles R: = 8.31 J/mol*K T: Temperature

Physics 218 Fall 2010 Example One mole (1.00 mol) of 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 nRT 1.00mol * 8.315 *300K V   molgK P 1.013x105 Pa 2 3 V  2.46x10 m

One cubic mm = 10-9 m3 V ' 1.0x109 m3 N  N  *6.02x1023 atoms  2.45x1016 atoms V A 2.46x102 m3 Physics 218 Fall 2010 Ideal Gas Law PV  nRT

For Constant Number of molecules: PV P V 1 1  2 2 T1 T2 For Constant Number of molecules & Temperature

Boyle’s Law

P1V1  P2V2

Physics 218 Fall 2010 Ideal Gas Law

For Constant n & T

Boyle’s Law

P1V1  P2V2

http://charles-boyle-law.org.ru/im/im5.jpg Physics 218 Fall 2010 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.

PV P V 1 1  nR  const.  2 2 T1 T2 P P 1  2 T1 T2

P2T1 552kPa *288K T2    315K P1 505kPa

Physics 218 Fall 2010

Kinetic of a molecule 3  2 BTkKE vm 2 of a molecule KE  2 3 Tk Average velocity v  B m

Physics 218 Fall 2010 Maxwell’s Speed Distribution

3 Tk v  B m

Physics 218 Fall 2010 Molecular Speed:

3kT v  rms m

Escape Speed:

2GM E ve   11,200m / s RE

Physics 218 Fall 2010 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 C. is 2 times the original speed D. is 2 times the original speed

Physics 218 Fall 2010