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5/3/16 Through out the PowerPoint lecture on specific heat there where 10 questions. Follow the PowerPoint lecture to answer the questions. The 10 questions.

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Presentation on theme: "5/3/16 Through out the PowerPoint lecture on specific heat there where 10 questions. Follow the PowerPoint lecture to answer the questions. The 10 questions."— Presentation transcript:

1 5/3/16 Through out the PowerPoint lecture on specific heat there where 10 questions. Follow the PowerPoint lecture to answer the questions. The 10 questions were collected at the end of the classroom for points

2 Warm UP 1) A 1000 Kg car is moving at 25 m/s as it exits a free-way. The driver brakes to meet the speed limit of 10 m/s. (a)What was the car’s kinetic energy on the freeway? (b) What was the work done by the car brakes and other friction forces in the process? How much?

3 cC calorie vs Calorie 1 Calorie = 1,000 calories

4 Definition of calorie lower case calorie vs Upper case Calorie The word “Calorie” — with a capital “C” — means 1000 calories (with a lowercase “c”), also known as a “kilocalorie” (also with a lowercase “c”). A calorie is defined as the amount of energy needed to raise the temperature of one gram of water one degree Celsius 1 calorie = 4.184 J in physics use ( Joules )

5 Energy content of food

6 the other unit used to measure heat is the Joule. The joule is the SI (System International) unit. In this juice label from Amsterdam you can see that energy (“energie”) is given in kilojoules and kilocalories.

7 Joule’s most famous experimental setups for demonstrating the mechanical equivalent of heat (PE) (work done by the gravitational force) converted into kinetic energy (KE) turbulence into increased average kinetic energy of atoms Resulting in temperature increase

8 Joule’s most famous experiment 2) If total mass of 11.5 kg falls 1.3 meters and all of the mechanical energy is converted to internal energy, by now much will internal energy of the water increase?

9 3) How much heat energy ( Joules) is required to increase the temperature of 1 gram of water by 1°C? It would take 4.184 J Different materials would warm up at different rates because each material has its own specific heat capacity Specific heat capacity refers to the amount of heat required to cause a unit of mass (say a gram or a kilogram) to change its temperature by 1°C

10 Essential question: How does one measure the quantity of heat gained or released by an object? Specific Heat Capacity– conceptual

11 Suppose that several objects composed of different materials are heated in the same manner. Will the objects warm up at equal rates? Oil in pan Water

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13 4) How much heat energy ( Joules) is required to increase the temperature of 2 gram of water by 1°C ? 5) How much heat energy ( Joules) is required to increase the temperature of 1 gram of water by 2°C ? The answer for both questions is (2 x4.184 J)

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15 Specific (c) heat in grams and Kg

16 6 A -Which material has the highest specific heat “c”? B- Which material has the lowest specific heat “c”?

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18 Why is water commonly used in the cooling system of a motor car?

19 451°F = 232 °C Energy for combustion of paper

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21 THE USE OF WATER IN FIRE EXTINGUISHING 7) Explain water’s contribution in putting out fires?_____________

22 Big IDEA

23 THE USE OF WATER IN FIRE EXTINGUISHING what is water’s contribution in putting out fires? Water vaporizes Water vapor occupies large volume, takes air out of the blaze Contributes to the fire’s cooling since water vapor absorbs heat ( high specific heat of the water) Contributes to the fire’s cooling since water vapor absorbs heat ( high specific heat of the water)

24 Don't use water to put out a grease fire https://www.youtube.com/watch?v=NC45vRE oEOg https://www.youtube.com/watch?v=NC45vRE oEOg

25 Q = mcΔT Q = mc (T f –T i ) i 8) What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? ( The specific heat capacity of water c = 4.18 J/g/°C. ) Show your work

26 Q = mCΔT What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C. Givens: m = 450 g c = 4.18 J/g/°C T initial = 15°C T final = 85°C Unknown Q = ??? ΔT = T final - T initial = 85°C - 15°C = 70.°C Q = mCΔT = (450 g)(4.18 J/g/°C)(70.°C) Q = 131670 J Q = 1.3x10 5

27 Relating the Quantity of Heat to the Temperature Change

28 9) What is change in temperature of 5 kg of water if it is given 84,000 J of heat energy? specific heat capacity of water = 4,200 J/(kg °C )

29 What is the rise in temperature of 5 kg of water if it is given 84,000 J of heat energy? specific heat capacity of water = 4,200 J/(kg K) Q = mcΔT Given: Q = 84,000 J m = 5 kg c = 4,200 J/(kg °C ) Looking for ΔT =? ΔT = Q / mc 84,000 J / ( 5kg x 4,200 J/kg °C) = 4 °C

30 Calorimetry –is a technique to measure transfer of thermal energy ( heat flow )

31 Specific heat and the transfer of objects between two objects Use Conservation of energy Lost of Heat by an object = Gained in heat by the second objects | Q Lost | = | Q Gained | mc ( T i – T f ) = mc (T f –T i ) Final Temperature is the same ( the two objects are in thermal equilibrium)

32 Calorimetry – finding the specific heat of a 1kg horseshoe 10) A 1kg horseshoe has average kinetic energy of 200°C. The horseshoe is placed in 5 liter (5kg) bucket of water. Initially, the water has a temperature of 20°C.The horseshoe and the water reach a thermal equilibrium of 21.8 °C. Find the specific heat of the horseshoe. Reach equilibrium

33 Given: Horseshoe m h = 1kg T ih = 200°C T fh = 21.8°C C h =???? Water m w = 5kg T iw = 20 °C T fw = 21.8°C c = 4,186 J/kg °C

34 Two ways to solve Isolate C h algebraically mc w ( T i – T f ) = m c h (T f –T i ) mc w ( T i – T f ) / (m h (T f –T i ))=c Q qained water = Q lost horseshoe Solve for Q gained by water Q = mc w ( T i – T f ) Then c h = Q / ( m w ( T ih – T f ) 211. 4 J

35 Conservation of energy Lost of Heat by an object = Gained in heat by the second object | Q Lost | = | Q Gained | mc ( T i – T f ) = mc (T f –T i ) Final Temperature is the same ( the two objects are in thermal equilibrium) A 60 kg of aluminum that is at 15 °C is placed into 20 kg of 90 °C water. Find the final temperature Specific heat of aluminum is c aluminum = 900 J/ kg °C Specific heat of water c water = 4,186 J/kg °C )


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