Part One Heat and Temperature
What does it mean to have a temperature of 0 C? What is temperature? Is temperature the same thing as heat?
Temperature is a measure of how “hot” or “cold” something is. Temperature is measured in arbitrary units, like Fahrenheit or Celsius.
Temperature is proportional to the average kinetic energy of the molecules of the substance.
Heat is the thermal energy transferred from a hot object to a cold object. Heat is measured in energy units -- Joules or calories.
Heat has the symbol q and is calculated using … q = mcDT
q = mcDT mass specific heat capacity Quantity of heat temperature change
The specific heat capacity of water is 4.18 J/gC Quantity of heat q = mcDT The specific heat capacity of water is 4.18 J/gC
q = m c DT q = (25.6g)(4.18J/gC)(30.0C) q = 3210 J How much heat is needed to raise the temperature of 25.6 grams of water from 20.0 C to 50.0 C? q = m c DT q = (25.6g)(4.18J/gC)(30.0C) q = 3210 J
q = m c DT = q DT m c Answer: 6.20 C What is the final temperature of 27.0 grams of liquid water, initially at 0C, after it absorbs 700.0 J of energy? q = m c DT Hint: start by solving for DT. = q DT m c Answer: 6.20 C
Part Two Phase Changes
We now know that heat is either absorbed or released during a phase change.
A process that absorbs heat is called endothermic. A process that gives off heat is called exothermic.
Endothermic: Exothermic: Melting (fusion) Vaporization Sublimation Heat is absorbed. Freezing Condensation Deposition Exothermic: Heat is released.
Heat is absorbed by the ice. And melts.
Heat is absorbed by the ice. One gram of ice at 0C absorbs 334 J as it melts to form water at 0C. … making liquid water
Heat is released by the water as it freezes. 334 joules is released when one gram of water freezes at 0C. Ice water
Ice Water releases 334 J per gram as it freezes at 0C Ice absorbs 334 J per gram as it melts at 0C Ice Water releases 334 J per gram as it freezes at 0C
Steam releases 2260 J/g as it condenses at 100 C Water absorbs 2260 J/g as it boils at 100 C Hotplate
The heat gained or lost in phase changes can be calculated using … Heat of fusion (melting) q = mHf q = mHv Heat of vaporization
Hf = 334 J/g Hv=2260 J/g The values for water are … Heat of fusion (melting) Hf = 334 J/g Hv=2260 J/g Heat of vaporization
q = m Hf q = (150.0 g)(334 J/g) q = 50,100 J or 50.1 kJ How much heat is absorbed by 150.0 g of ice as it melts at 0C? q = m Hf q = (150.0 g)(334 J/g) q = 50,100 J or 50.1 kJ
q = m Hv q = (20.0 g)(2260 J/g) q = 45,200 J or 45.2 kJ How much heat is released by 20.0 grams of steam as it condenses at 100C? q = m Hv q = (20.0 g)(2260 J/g) q = 45,200 J or 45.2 kJ
Part Three Phase Diagrams
2 1 3 Which phase is in each region? Hint: What happens to ice as temperature increases? The phase diagram has three distinct regions. Temperature Pressure 2 1 3
The point where all three phases exist in equilibrium is called the triple point. triple point. L S Pressure G Temperature
At a pressure of 1 atm, most substances go through all three phases, as the temperature increases, Solids melt to form liquids, which vaporize to form gases. 1 atm G Temp.
Temp. S L G 1 atm MP BP Notice the melting point and boiling point.
L S G But the phase diagram for CO2 is a little different. Notice that the triple point is above 1 atm. L S 5 atm 1 atm G Temperature
At 1 atm CO2 goes directly from solid to vapor as the temperature increases. The sublimation point is –78.5 C 1 atm G Temperature
L S G What phase change is occurring? Melting (fusion) Pressure Temperature
L S G What phase change is occurring? Vaporization Pressure Temperature
L S G What phase change is occurring? Condensation Pressure Temperature
What phase change is occurring? L G Sublimation Pressure Temperature
L S G What phase change is occurring? Liquefying a gas by increasing the pressure. S L G Pressure Temperature
Heating and Cooling Curves Part Four Heating and Cooling Curves
Look at the different regions of the heating curve for water. 100 Time Temp Water and steam Steam Water Ice and water Phase changes? Ice
Calorimetry and Specific Heat Capacity Part Five Calorimetry and Specific Heat Capacity
Calorimetry is a collection of laboratory procedures used to investigate the transfer of heat. In calorimetry experiments, one might be looking for a final temperature or a specific heat capacity.
What is the law of conservation of energy? Energy is neither created nor destroyed, only changed in form.
The law of conservation of energy suggests that the heat lost by the hot object as it cools is equal to the heat gained by the cool water as it warms up.
qlost = -qgained = mocDTo = -mwcDTw Investigate: To put it mathematically: qlost = -qgained Heat lost by the hot object Heat gained by the cold water = And since q = mcDT then mocDTo = -mwcDTw
moc(Tf -Ti) = -mwc(Tf -Ti) Investigate: The convention for DT is final temperature minus initial temperature or Tfinal – Tinitial mhcDTo = -mccDTw becomes moc(Tf -Ti) = -mwc(Tf -Ti) Use your algebra skills, to solve for Tf , the final temperature.
Specific heat capacity … …varies from one substance to another. …a measure of how much heat something can “hold”. …the amount of heat needed to raise one gram of a substance by one Celsius degree.
Specific heat capacity lab suggestions: Heat a metal to a known temp. Transfer the metal to a known quantity of water at a known temperature. Measure the equilibrium temperature. Use qlost = -qgained to compute the specific heat of the metal.
Get the initial temperature of the metal. The temperature of boiling water. metal hotplate
Get initial temp of water in calorimeter cup.
Transfer the metal to the calorimeter.
Continue stirring.
Data: Mass of metal Initial temp of metal Mass of water Initial temp of water Final temp of water and metal
qlost = -qgained mmcmDTm = -mwcwDTw -mwcwDTw cm = mmDTm
Mass of metal 40.0 g Initial T of metal 98.0 C Mass of water in calorimeter 60.0 g Initial T of water 20.0 C Final T of water and metal 22.9 C Calculate the specific heat capacity of the metal.
Table of selected specific heats. What is the unknown metal?