chapter 6 notes heat heat & temperature calculations

Chapter 6 NotesHEAT

Heat & Temperature Calculations

Temperature = a measure of the AVERAGE kinetic energy in the substance.

Celsius (°C)Fahrenheit

(°F)Kelvin (°K)

NEED TO FIND FORMULA

°F 1.8°C + 32

°C °F – 32/1.8

°K °C + 273

°C °K – 273

°F °K °C °F

0°K = absolute zero = all molecular motion stops

H20 distilled water (pure water)

melting point = 0°Cboiling point = 100°C

Melting Points examples

Gallium a# 31 M.P. 86oF

Iron a# 26 M.P. 2800oF

Mercury a# 80 M.P. -38oF

Gold a# 79 M.P. 1947oF

Copper a# 29 M.P. 1984oF

Boiling Pointsexamples

Gallium a# 31 B.P. 3999oF

Iron a# 26 B.P. 5182oF

Mercury a# 80 B.P. 674oF

Gold a# 79 B.P. 5173oF

Copper a# 29 B.P. 4644oF

Energy (heat) measure in Joules, BTUs (British Thermal Units) calories and Calories.

1 calories = 4.186 Joules1 BTU = 252 calories1 Calorie = 1000 calories

States of Matter Also called Phases of Matter

Solids

Liquids

Vapors (gases)

Solids

Have a definite shape

Have a definite volume

Particles VIBRATE in place

Liquids

Have NO definite shape

Have definite volume

particles SLIDE freely

Gases (vapor)

Have NO definite shape

Have NO definite volume

particles fill the volume of the container

Solids, Liquids & Gases

Solids = can form crystals = solid where the particle are arranged into repeating patterns.Liquids = physical property of Viscosity = “thickness” – the resistance to flow.Gases = volume of gases depend greatly on pressure and temperature.

Phase Changes

MeltingFreezingVaporizationCondensationSublimationphysical changes

Melting

the process of changing from a solid to a liquid

energy is being put into the substance

melting point = the temperature at which melting occurs – physical property

the melting point of water is 0ºC

Freezing

the process of changing from a liquid to a solid energy is being pulled out of the substancefreezing point = same temperature as the melting point (used mainly in weather)

Vaporization

the process of changing from a liquid to a gas

energy in being put into the substance

evaporation

boiling

Evaporation

vaporization that occurs at the surface of the liquid

Boiling

vaporization that occurs throughout the liquidboiling point = the temperature at which boiling occursthe boiling point of water is 100ºC

Condensation

the process of changing from a gas to a liquid

energy is being pulled out of the substance

Sublimation

the process of changing from a solid to a gas

energy is being put into the substance

ex: dry ice (CO2)

heat of fusion

Heating of water

0°C

100°C

heat of vaporization

ICE

WATER (liquid)

STEAM

Heat Transfer

Conduction

Convection

Radiation

Conduction transfer of heat by direct contact

(molecule to molecule)metals are good conductorspoor conductors = insulators

Convection transfer of heat by “convection currents”

warm fluids are less dense than colder fluid thus warm fluids rise and cold fall.

not possible in solids fluid = anything that flows (liquids &

gases) hot air balloons, “convection” ovens

Radiation transfer of heat by electromagnetic

wavessome wavelengths of infrared &

ultravioletonly type of transfer that can occur

through empty spacesun Earth

Specific Heat

The amount of heat needed to raise the temperature of one gram of a substance one degree Celsius.

Factors in Specific Heat

types of substance (C)

mass of the substance (m)

how much of a temperature change (∆T)

C = specific heat constant

m = mass∆T = difference

in the temperature

Specific Heat Calculations

∆Q = amount of heat absorbed (difference in the heat or heat change)

∆Q = m x ∆T x C

The specific heat of water= 1.0 cal/g°C or = 4.2 joules/ g°C

EXAMPLE #1:How many calories are absorbed by a pot of water with a mass of 500 grams in order to raise the temperature from 20°C to 30°C?

C = 1.0 cal/g°Cm = 500 grams∆T = 10°C (30-

20)

∆Q = m x ∆T x C ∆Q = (500 g)(10°C)(1.0 cal/g°C) ∆Q = 5000

calories

EXAMPLE #2:How many joules are absorbed by a pot of water with a mass of 500 grams in order to raise the temperature from 20°C to 30°C?

C = 4.2 J/g°Cm = 500 grams∆T = 10°C (30-

20)

∆Q = m x ∆T x C ∆Q = (500 g)(10°C)(4.2 J/g°C) ∆Q = 21,000

Joules

Phase Changes

Heat of fusion (Hf) the heat energy needed to melt (or

freeze) a substance. All heat being put into the substance

goes to the melting process thus the temperature does not

change while the substance is melting.

Phase Changes

Heat of vaporization (Hv) the heat energy needed to boil (or

condense) a substance. All heat being put into the substance

goes to the boiling process thus the temperature does not

change while the substance is boiling.

Heat & Phase Changes

Hf = mass x Hf constantThe heat of fusion of water = 340 J/gHv = mass x Hv constantThe heat of vaporization of water = 2300

J/g

EXAMPLE:How many joules of heat are necessary to melt 500 g of ice?

Chf = 340 J/gm = 500 gH = Chf x mH = (340 J/g)(500 g)H = 170,000 J