Temperature & Matter II MC Textbook Chp 9, 11 GLM Unit 11

Topics 1) Thermal Physics of Gases 2) Specific Heat Capacity & Heat Capacity 3) Specific Latent Heat & Latent Heat

Part 1: Thermal physics of gases

Recall: Kinetic Model of Gases Particles are randomly arranged Particles are far apart from each other (negligible attractive forces between them) Particles move randomly at fast speeds Upon heating, particles move faster

Microscopic Gas Pressure When a gas is held in a container, the collisions between the particles and the container give rise to gas pressure 2 factors to gas pressure: The greater the rate of collisions, the greater the pressure The faster each molecule moves, the more forceful the collision against the container, the greater the pressure

3 Ways to Increase Pressure of Gas Reduce the volume of the container (i.e. compress the gas) Increases the rate of collisions between gas particles and container Pump more gas into the container Heat the gas Increase speed of gas molecules, collisions are more forceful

Gas Laws For gases, there are mainly 3 macroscopic properties we are interested in: Temperature, Volume and Pressure The gas laws relate the relationships between these three quantities In each gas law, one of the quantity is held constant Note that the gas laws only apply for a fixed mass of gas (they are invalid when you try to add more gas or remove some gas)

Gas Law 1: Fixed Volume Possible situation: gas is trapped in a rigid container and heated/cooled Temperature (in Kelvin) is directly proportional to Pressure E.g. when temperature is doubled, pressure is doubled E.g. when temperature is halved, pressure is halved

Gas Law 2: Fixed Pressure Possible Situation: Air in an elastic balloon is heated up/cooled down Temperature (in Kelvin) is directly proportional to volume E.g. When temperature is doubled, volume is doubles E.g. When temperature is halved, volume is halved

Gas Law 3: Fixed Temperature Possible Situation: A balloon of air is submerged into water/taken out of water Pressure is inversely proportional to volume E.g. when pressure is doubled, volume is halved E.g. when pressure is halved, volume is doubled

Practice Task GLM Pg 138 Qn 1

part 2: Specific heat capacity

Introduction In this topic we will study the relationship between heat (which is energy) and temperature (i.e. how hot or how cold a substance is) Suppose I use a bunsen burner to heat up 1 kg piece of metal, and I use the same bunsen burner to heat up 1 kg of water. Which one would heat up to 100°C faster?? Ans: the metal. Why? The metal requires less heat (thermal energy) to increase its temperature, while water requires more heat to increase the same amount of temperature.

Specific Heat Capacity We say that water has a higher specific heat capacity than metal Specific heat capacity is defined as the amount of thermal energy required to raise the temperature of a unit mass of a substance by 1 K or 1° Symbol: small case “c” Units: Jkg-1K-1 OR Jg-1K-1

Specific Heat Capacity Analogy: Dawn only need to eat 1 plate of rice to feel full, Zorye needs to eat 2.5 plates of rice to feel equally full. After they both have eaten, even though they are equally fully, Zorye needs 2.5x as much food to feel just as full. Similarly, if material X has twice as much specific heat capacity as material Y, 1 kg of X would require twice as much thermal energy (heat) to increase its temperature by the same amount compared to 1 kg of Y

Specific Heat Capacity Equation to memorize: Q = mcθ Q is heat supplied or given out, in Joules m is mass of substance, in kg (or g) c is specific heat capacity, in Jkg-1K-1 (or Jg-1K-1) θ is change in temperature, in °C or K Protip: use this formula to remember units of c

Example 1 Water has specific heat capacity of 4200 Jkg-1K-1 . If 5000 J of heat was supplied to 1 kg of water, determine the increase in temperature of the water. Q = mcθ 5000 = (1)(4200)θ θ = 1.19 °C (3 sf) The water increased its temperature by 1.19 °C

Example 2a A cup of containing 200g of boiling water was allowed to cool from 100°C to 25°C. If the specific heat capacity of water is 4200 Jkg-1K-1, determine how much heat was given out by the water if the water is in thermal equilibrium with the cup.

Example 2b If the mass of the cup is 300 g and the specific heat capacity of the cup is 2.00 Jg-1K-1 , determine the total amount of heat given off by the cup AND water.

Practice Task GLM Pg 177 Qn 6(b)

Strategy for solving more complex problems When there are two or more substances interacting (i.e, passing heat from one to another), Step 1: treat the two substances separately in your working to determine Q for each substance Step 2: Heat gain by one substance = heat loss by other substance If temperature is an unknown, use algebra to solve (e.g. Let X be the final Temperature)

Example 3 200 g of cold water at 10 °C is mixed with 300g of warm water at 50 °C. What is the resulting temperature of water, if the specific heat capacity of water is 4200 Jkg-1K-1 ? Assume no heat loss to surroundings.

Example 4 A 150 g piece of metal was heated to 200 °C, and then placed in 500 g of water, initially at 25 °C. Determine the final temperature when the metal and water are in thermal equilibrium. Assume no heat loss to surroundings. The Specific heat capacities of metal and water are 1000 Jkg-1K-1 and 4200 Jkg-1K-1 respectively.

Practice Task GLM Pg 176 Qn 2 GLM Pg 178 Qn 6(d)

Heat Capacity Heat Capacity Symbol: capital letter “C” Units: JK-1 not to be confused with specific heat capacity Symbol: capital letter “C” Units: JK-1 relationship to c: C = mc m is mass c is specific heat capacity Definition: Heat Capacity C is the amount of thermal energy required to raise the temperature of a substance by 1 K (or 1 C)

Example 5 The specific heat capacities of water and ethanol are 4200 Jkg-1K-1 and 2450 Jkg-1K-1. Beaker A contains 200g of Water, and Beaker B contains 350g of ethanol. Determine their relative heat capacities, and hence state which beaker would require more heat to increase the temperature by 1 K.

part 3: Specific latent heat

Specific Latent Heat Just as specific heat capacity is the heat required to increase the temperature of 1 kg of substance, specific latent heat is the heat required to change the state of 1 kg of substance Symbol: lower case l Units: J kg-1 Equation to memorize: Q = ml

Specific Latent Heat There are two kinds of specific latent heat. Specific latent heat of fusion lf of a substance is the amount of thermal energy required to change unit mass of the substance from solid state to liquid state, without a change in temperature Specific latent heat of vaporisation lvof a substance is the thermal energy required to change unit mass of the substance from liquid state to gaseous state, without a change in temperature

Example 6 The specific latent heat of vaporisation of water is 2200 kJ kg-1. Determine how much water is converted to steam when 1000 kJ of energy is supplied to water at 100 °C.

Example 7 51 kJ of energy is released when 150 g of substance X solidifies at melting point, determine the specific latent heat of fusion of X.

Practice Task GLM Pg 180 Qn 4, 6

Latent Heat Latent Heat Symbol: Capital letter “L” Units: J Not to be confused with specific latent heat. Symbol: Capital letter “L” Units: J Relationship to l: L = ml

Example 8 The specific latent heats of vapourization of alcohol and water are 855 kJkg-1 and 2260 kJkg-1 respectively. Determine which requires more heat to completely convert to gas at their boiling points – 500 g of water or 1.4 kg of alcohol.

Latent Heat Latent heat of fusion Lf is the amount of thermal energy required to change a substance from solid state to liquid state, without a change in temperature Latent heat of vaporisation Lv is the amount of thermal energy required to change a substance from liquid state to gaseous state, without a change in temperature.

Summary of 4 quantities Heat required per unit mass Heat required To Increase temperature by 1 °C Specific Heat Capacity Heat Capacity To change state Specific Latent Heat Latent Heat

Strategy for solving combined problems If there is more than one substance interacting, consider them separately. If one substance undergoes more than one change, break it apart into different phases and consider them separately (e.g. Phase 1 – increase of temperature to boiling point, Phase 2 – conversion from liquid to gas) Use algebra to solve for unknowns (e.g. let X be final temperature, etc.)

Worked Example 9 Specific heat capacity of water is 4200 Jkg-1K-1 and the specific latent heat of vaporisation is 2200 kJ kg-1. Determine how much energy it would need to convert 500g of water at 25 °C completely to steam.

Worked Example 10 Specific heat capacity of water is 4200 Jkg-1K-1 and the specific latent heat of vaporisation is 2200 kJ kg-1. 50 g of steam at 100 °C was pumped into 500g of water at 25 °C. If all the steam was condensed into water, determine the final temperature of water. Ans: 71.3 °C

Summary 6 definitions: Specific Heat Capacity, Heat Capacity, Specific Latent Heat, Latent Heat 2 Equations Q = mcθ Q = ml Solving quantitative problems which include either or both of these two equations

Quiz 11

Assignment 11 Topic 11A Paper 1 – Qn 2,3,4, 6 Paper 2 – Qn 1 Topic 11B