HeatHeat

 When two objects at different temperatures are put into contact, heat spontaneously flows from the hotter to the cooler one. If kept in contact long enough- Their temperatures will become equal  We call this thermal equilibrium  When two objects at different temperatures are put into contact, heat spontaneously flows from the hotter to the cooler one. If kept in contact long enough- Their temperatures will become equal  We call this thermal equilibrium

HEAT  Heat: the energy that is transferred from one body to another because of a difference in temperature.  Symbol for Heat: Q  SI unit for heat: (same as for energy)- Joule (J)  Heat: the energy that is transferred from one body to another because of a difference in temperature.  Symbol for Heat: Q  SI unit for heat: (same as for energy)- Joule (J)

 18 th century model had heat flow like a fluid- they called it caloric. Due to history we have various units for heat. CALORIE: (cal) – Note the lowercase ‘c’  This is the amount of heat needed to raise the temperature of 1g of H 2 O by 1°C  18 th century model had heat flow like a fluid- they called it caloric. Due to history we have various units for heat. CALORIE: (cal) – Note the lowercase ‘c’  This is the amount of heat needed to raise the temperature of 1g of H 2 O by 1°C

Kilocalorie: (kcal) this is 1000 calories  Heat needed to raise 1kg of water by 1°C Calorie Note the capital ‘C’  a Calorie is 1kcal, it is also called a dietary calorie Kilocalorie: (kcal) this is 1000 calories  Heat needed to raise 1kg of water by 1°C Calorie Note the capital ‘C’  a Calorie is 1kcal, it is also called a dietary calorie

British Thermal Units (BTU’s)  1 BTU=.252 kcal = 1055 J  Heat needed to raise temperature of 1lb of H 2 O by 1°F British Thermal Units (BTU’s)  1 BTU=.252 kcal = 1055 J  Heat needed to raise temperature of 1lb of H 2 O by 1°F

James Joule Experiment  The break through idea from this experiment is that work done had an equivalent amount of heat transferred. “Mechanical Equivalent of Heat”  Doing work (fd, mgh, ½mv 2 ) on an object raises its temperature 1 cal= J 1 kcal = 4186 J  The break through idea from this experiment is that work done had an equivalent amount of heat transferred. “Mechanical Equivalent of Heat”  Doing work (fd, mgh, ½mv 2 ) on an object raises its temperature 1 cal= J 1 kcal = 4186 J

Joule’s Experiment

Cake & Ice Cream= mgh Example  How many stairs would you need to climb to “work off” 500 Calories of cake and ice cream? Assume mass of person = 60kg 500 Calories = 500kcal 500 kcal (4186 J/kcal) = 2.1 x 10 6 J Work Done in climbing stairs = mgh 2.1 x 10 6 J = 60kg (9.8)h 3600m = h About 2 miles note- this assumes body is 100% efficient machine

Example  A 3g bullet is shot at a tree and passes through it. The bullet speed changes from 400m/s to 200m/s. How much heat (Q) is produces and shared by the bullet and tree. Q = ΔKE Q = KE f -KE i Q = ½m (V f 2 – V i 2 ) Q = ½ (.003kg)( ) Q= 180 J 180/4.186 = 43 cal Q = ΔKE Q = KE f -KE i Q = ½m (V f 2 – V i 2 ) Q = ½ (.003kg)( ) Q= 180 J 180/4.186 = 43 cal

Temperature, Heat, Internal Energy  The sum total of all the energy of all the molecules in an object is called: THERMAL ENERGY or INTERNAL ENERGY

Distinction between Temperature, Heat & Internal Energy  Temperature: (K) is the measure of average KE of individual molecules.  Thermal Energy: refers to the total energy of all the molecules in an object. (2 equal- mass chunks of Fe may have the same temperature, but the both of them have twice as much thermal energy as 1 does)  Heat: refers to a transfer of energy (thermal energy) from one object to another because of a difference in temperature.  Temperature: (K) is the measure of average KE of individual molecules.  Thermal Energy: refers to the total energy of all the molecules in an object. (2 equal- mass chunks of Fe may have the same temperature, but the both of them have twice as much thermal energy as 1 does)  Heat: refers to a transfer of energy (thermal energy) from one object to another because of a difference in temperature.

Specific Heat  The amount of heat required to change the temperature of a given material is proportional to the mass of the material and to the temperature change. Q = m c ΔT c= specific heat capacity c = Q/ mΔT Units- J/kg°C Table 14-1 on page 421 List of common specific heats  The amount of heat required to change the temperature of a given material is proportional to the mass of the material and to the temperature change. Q = m c ΔT c= specific heat capacity c = Q/ mΔT Units- J/kg°C Table 14-1 on page 421 List of common specific heats

Substance Aluminum Copper Glass Iron/Steel Lead Marble Silver Wood

Example  How much heat is required to raise the temperature of an empty 20kg vat made of iron from 10°C to 90°C? What if the vat is filled with 20kg of water? ΔT = 80°C  How much heat is required to raise the temperature of an empty 20kg vat made of iron from 10°C to 90°C? What if the vat is filled with 20kg of water? ΔT = 80°C Q = mc ΔT 20kg(450)80 = 7.2x10 5 J 720 KJ Q = mc ΔT 20(4186)80 = 6.7x 10 6 J 6700 KJ Total heat= 720 KJ KJ = 7400KJ Almost 10x more heat for equal amount of Iron Total heat= 720 KJ KJ = 7400KJ Almost 10x more heat for equal amount of Iron

Calorimetry  The quantitative measurement of heat exchange  The main idea is conservation of Energy  Heat Lost = Heat Gained  Q lost = Q gained  Heat”flows” from region of hotter to region of cooler until thermal equilibrium is reached  The quantitative measurement of heat exchange  The main idea is conservation of Energy  Heat Lost = Heat Gained  Q lost = Q gained  Heat”flows” from region of hotter to region of cooler until thermal equilibrium is reached

ExampleExample  If 200cm 3 of tea at 95°C is poured into a 150g glass coffee cup at 25°C, what will be the final temperature (T) of both when equilibrium is reached? (assume NO heat is lost to surroundings) Tea ~ Water 1cm 3 = 1ml = 1g (for H 2 O) 200cm 3 =.2kg Q lost tea = Q gained cup M T C T (95°C-T) = M c C c (T-25°C) T is final temp for both tea and cup

continued….2kg (4186)(95-T) = (.15kg)(840)(T-25°C) 79,400J – (836T) = 126T – 6300J T = 89°C Tea drops 6°C cup rises 64°C

 Note ΔT is Positive  Heat lost = T i – T f  Heat gained = T f – T i  Calorimeter: tool used in calorimetry  Note ΔT is Positive  Heat lost = T i – T f  Heat gained = T f – T i  Calorimeter: tool used in calorimetry

 It’s a cup within a cup, It insulates and allows for temperature measuring Air insulator Water

ExampleExample  A mystery alloy has been discovered in Lacey’s computer room. Find the specific heat capacity. The alloy sample is.15kg and is heated to 540°C. It is quickly placed in 400g of water at 10°C. The calorimeter cup is made of aluminum and has a mass of 200g. The final temp is 30.5°C

Q lost alloy = Q gained by water + Q gained by cup M A C A ΔT A = M W C W ΔT W + M C C C ΔT C.15(C A )( ) =.4(4186)( ) +.2(900)(30.5 – 10).15(C A )( ) =.4(4186)( ) +.2(900)(30.5 – 10) 76.4C A = (34, ,700) J/kg 0 C C A = 500 J/kg 0 C

Latent Heat  Deals with Q during a change of phase ; solid to liquid, liquid to gas  Latent means hidden  L F : latent heat of fusion  L V : latent heat of Vaporization Table 14-3 Pg. 425

ExampleExample  How much Q does a refrigerator have to remove from 1.5kg of water at 20°C to make ice at –12°C? Q Q total = Q 1 + Q 2 + Q 3

Q total = MC W ΔT + ML F + MC ice ΔT = 1.5(4186)(20) + 1.5(3.33 x 10 5 ) + 1.5(2100)12 = 1.5(4186)(20) + 1.5(3.33 x 10 5 ) + 1.5(2100) x 10 5 J 660kJ Read example 14-9 and on pg

 Conduction 14-7  Convection 14-8  Radiation 14-9  Conduction 14-7  Convection 14-8  Radiation 14-9