1 PHYS1001 Physics 1 REGULAR Module 2 Thermal Physics HEAT CAPACITY LATENT HEAT What is cooking all about? ptC_heat.ppt

2 Overview of Thermal Physics Module: 1.Thermodynamic Systems: Work, Heat, Internal Energy 0 th, 1 st and 2 nd Law of Thermodynamics 2.Thermal Expansion 3.Heat Capacity, Latent Heat 4.Methods of Heat Transfer: Conduction, Convection, Radiation 5.Ideal Gases, Kinetic Theory Model 6.Second Law of Thermodynamics Entropy and Disorder 7.Heat Engines, Refrigerators

3  HEAT CAPACITY LATENT HEAT §17.5 p582 Heat Heat capacity (specific heat capacity) Phase changes Conservation of energy calorimetry References: University Physics 12 th ed Young & Freedman

4 What happens when we heat a substance? How does the temperature change? When does the state of matter change (phase changes)?

5 Phases of matter Gas - very weak intermolecular forces, rapid random motion Liquid - intermolecular forces bind closest neighbours Solid - strong intermolecular forces low temp high pressure high temp low pressure

6 P = 2000 WTime t (min) Heat Q (kJ) Time Ratios Ice: -20 o C to 0 o C (0.37/0.37) Ice / Water: 0 o C (2.78/0.37) Water: 0 o C to 20 o C Water: 0 o C to 100 o C Water / Steam: 100 o C Steam: 100 o C to 120 o C Heating 1.0 kg ice to steam (-20 o C to 120 o C)

7 Q(t)Q(t) T(t) melting point boiling point s s / l l l /g g mc s  T mL f mc l  T mL v mc g  T Simple model for heating a substance at a constant rate

8 solid liquid gas sublimation melting freezing evaporation condensation Q= ± m L f Q= ± m L V At melting point: L f latent heat of fusion or heat of fusion At boiling point: L V latent heat of vaporization or heat of vaporization Q > 0 energy absorbed by substance during phase change Q < 0 energy released by substance during phase change  Phase changes at constant temperature Q= ± m L S deposition L S latent heat of sublimation or heat of sublimation

9 TiTi TfTf  T= T f - T i Mass of object m Specific heat (capacity) c  Specific heat or specific heat capacity, c NO phase change during temperature change heat Q

10 Specific heat Latent heats Substance c (J.kg -1.K -1 ) Aluminum 910 Copper 390 Ice 2100 Water 4190 Steam 2010 Air 1000 Soils / sand ~500 Latent heat – phase change (formation or breakage of chemical bonds requires or releases energy) Water - large values of latent heats at atmospheric pressure L f = 3.34x10 5 J.kg -1 (273 K) L v = 2.26x10 6 J.kg -1 (373 K) You can be badly scolded by steam – more dangerous than an equivalent amount of boiling water WHY?

11 Latent heat and phase changes As a liquid evaporates it extracts energy from its surroundings and hence the surroundings are cooled. When a gas condenses energy is released into the surroundings. Steam heating systems are used in buildings. A boiler produces steam and energy is given out as the steam condenses in radiators located in rooms of the building.

12 Evaporation and cooling Evaporation rates increase with temperature, volatility of substance, area and lower humidity. You feel uncomfortable on hot humid days because perspiration on the skin surface does not evaporate and the body can't cool itself effectively. The circulation of air from a fan pushes water molecules away from the skin more rapidly helping evaporation and hence cooling. Evaporative cooling is used to cool buildings. Why do dogs pant? When ether is placed on the skin it evaporates so quickly that the skin feels frozen. Ethyl chloride when sprayed on the skin evaporates so rapidly the skin is "frozen" and local surgery can be performed.

13 Problem C.1 The energy released when water condenses during a thunderstorm can be very large. Calculate the energy released into the atmosphere for a typical small storm. Where did the water come from?

14 Solution Identify / Setup Assume 10 mm of rain falls over a circular area of radius 1 km h = 10 mm = m r = 1 km = 10 3 m volume of water V =  r 2 h =  (10 6 )(10 -2 ) = 3  10 4 m 3 mass of water m = ? kg density of water  = 10 3 kg.m -3 m =  V = (10 3 )(3  10 4 ) = 3  10 7 kg Latent heat – change of phase Q = m L L v = 2.26  10 6 J.kg -1 Energy released in atmosphere due to condensation of water vapour h r

15 Execute Q = m L V = (3  10 7 )(2.26  10 6 ) J = 7  J Evaluate The energy released into the atmosphere by condensation for a small thunder storm is more than 10 times greater then the energy released by one of the atomic bombs dropped on Japan in WW2. This calculation gives an indication of the enormous energy transformations that occur in atmospheric processes.

16 Problem C.2 For a 70 kg person (specific heat 3500 J.kg -1.K -1 ), how much extra released energy would be required to raise the temperature from 37 °C to 40 °C? Solution Identify / Setup m = 70 kg c = 3500 J.kg -1.K -1  T = (40 – 37) °C = 3 °C Specific heat capacity Q = m c  T Execute Q = m c  T = (70)(3500)(3) J = 7.4  10 5 J = 0.74 MJ Evaluate

17 Water has a very large specific heat capacity compared to other substances c water = 4190 J.kg -1.K -1 The large heat capacity of water makes it a good temperature regulator, since a great amount of energy is transferred for a given change in temperature. Why is there a bigger difference between the max and min daily temperatures at Campbelltown compared to Bondi? Why is water a good substance to use in a hot water bottle? Why is the high water content of our bodies ( c ~ 3500 J.kg -1.K -1 ) important in relation to the maintenance of a constant core body temperature?

18 CALORIMETRY PROBLEMS This is a very common type of problem based upon the conservation of energy. It involved changes in temperature and phase changes due to heat exchanges. Setup All quantities are taken as positive in this method. Identify the heat exchanges (gained or lost), phase changes and temperature changes. Conservation of energy energy gained = energy lost

19 Problem C.3 How much ice at –10.0 °C must be added to 4.00 kg of water at 20.0 °C to cause the resulting mixture to reach thermal equilibrium at 5.0 °C. Sketch two graphs showing the change in temperature of the ice and the temperature of the water as functions of time. Assume no energy transfer to the surrounding environment, so that energy transfer occurs only between the water and ice.

20 ICE gains energy from the water Ice -10 o C Ice/water 0 o CWater 5 o C Water 20 o CWater 5 o C WATER losses energy to the ice Heat gained by ice Q ice = heat lost by water Q water Solution Identify / Setup Q = m c  T Q = m L

21 m ice = ? kg m water = 4.00 kg temperature rise for ice to melt  T ice1 = 0 – (–10) °C = 10 °C temperature rise of melted ice  T ice2 = (5 – 0) °C = 5 ° C temperature fall for water  T water = (20 – 5) °C = 15 °C c ice = 2100 J.kg -1.K -1 c water = 4190 J.kg -1.K -1 L f = 3.34  10 5 J.kg -1 NB all temperature changes are positive

22 Execute T t T t water ice 0 o C equilibrium temperature reached

23 Q ice =m ice c ice  T ice1 + m ice L f + m ice c water  T ice2 Q water = m water c water  T water Q ice = Q water m ice c ice  T ice1 + m ice L f + m ice c water  T ice2 = m water c water  T water alternatively can calculate each term Evaluate heat gained by ice heat lost by water conservation of energy

24 Problem C.4 June 2007 Exam Question (5 mark) A sample of liquid water A and a sample of ice B of identical masses, are placed in a thermally isolated container and allowed to come to thermal equilibrium. The diagram below is a sketch of the temperature T of the samples verses time t. Answer each of the following questions and justify your answer in each case. 1 Is the equilibrium temperature above, below or at the freezing point of water? 2 Has the liquid water partly frozen, fully frozen, or not at all? 3 Does the ice partly melt, or does it undergo no melting? T t A B

25 Solution Identify / Setup Phase change – temperature remains constant Q =  m L Ice melts at 0 o C and liquid water freezes at 0 o C Temperature change – no change in phase Q = m c  T Ice warms and liquid water cools Energy lost by liquid water (drop in temperature) = Energy gained by ice (rise in temperature + phase change)

26 (1) Ice increase in temperature initially and then remains constant when there is a change in phase. Therefore, the equilibrium temperature reached is the freezing point. (2) The ice reaches the freezing point first and the then the temperature remains constant. As the water cools, the ice melts. The temperature never rises above the freezing point, therefore, only part of the ice melts. (3) The temperature of the water falls to its freezing point and never falls below this and hence it is most likely that no liquid freezes. Execute

27 Y&F Example What’s Cooking Cu pot m = 2.0 kg T = 150 o C Water added m = 0.10 kg T = 25 o C Final temperature of water and pot T f ? energy lost by pot = energy gained by water: 3 possible outcomes 1 none of water boils 25 o C < T f < 100 o C 2 some of the water boils T f = 100 o C 3 all water boils to steam 100 < T f < 150 o C have to becareful with such problems

28 How do we measure a person’s metabolic rate? Santorio Santorio weighed himself before and after a meal, conducting the first controlled test of metabolism, AD 1614.

29 A fever represents a large amount of extra energy released. The metabolic rate depends to a large extent on the temperature of the body. The rate of chemical reactions are very sensitive to temperature and even a small increase in the body's core temperature can increase the metabolic rate quite significantly. If there is an increase of about 1 °C then the metabolic rate can increase by as much as 10%. Therefore, an increase in core temperature of 3% can produce a 30% increase in metabolic rate. If the body's temperature drops by 3 °C the metabolic rate and oxygen consumption decrease by about 30%. This is why animals hibernating have a low body temperature. During heart operations, the person's temperature maybe lowered.