Heat & Temperature Calculations Chapter 6 Notes HEAT Heat & Temperature Calculations

Temperature = a measure of the AVERAGE kinetic energy in the substance. Celsius (°C) Fahrenheit (°F) Kelvin (°K)

0°K = absolute zero = all molecular motion stops NEED TO FIND FORMULA °F 1.8°C + 32 °C °F – 32/1.8 °K °C + 273 °K – 273 °K °C °F 0°K = absolute zero = all molecular motion stops

H20 distilled water (pure water) melting point = 0°C boiling 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 Points examples 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 Joules 1 BTU = 252 calories 1 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 Melting Freezing Vaporization Condensation Sublimation physical 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 substance freezing 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 liquid boiling point = the temperature at which boiling occurs the 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)

Heating of water heat of vaporization heat of fusion STEAM 100°C WATER (liquid) heat of vaporization 0°C ICE heat of fusion

Heat Transfer Conduction Convection Radiation

Conduction transfer of heat by direct contact (molecule to molecule) metals are good conductors poor 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 waves some wavelengths of infrared & ultraviolet only type of transfer that can occur through empty space sun  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°C m = 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? ∆Q = m x ∆T x C ∆Q = (500 g)(10°C)(4.2 J/g°C) ∆Q = 21,000 Joules C = 4.2 J/g°C m = 500 grams ∆T = 10°C (30-20)

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 constant The heat of fusion of water = 340 J/g Hv = mass x Hv constant The 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/g m = 500 g H = Chf x m H = (340 J/g)(500 g) H = 170,000 J