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Chapter 1 Temperature and heat. 1-1 Zeroth law of thermodynamics " If object A is in thermal equilibrium with object B, and object C is also in thermal.

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Presentation on theme: "Chapter 1 Temperature and heat. 1-1 Zeroth law of thermodynamics " If object A is in thermal equilibrium with object B, and object C is also in thermal."— Presentation transcript:

1 Chapter 1 Temperature and heat

2 1-1 Zeroth law of thermodynamics " If object A is in thermal equilibrium with object B, and object C is also in thermal equilibrium with object B, then objects A and C will be in thermal equilibrium if brought into thermal contact.

3 Continue: اذا كان هناك جسم A فى حالة اتزان حرارى مع جسم اخر B وجسم C فى حالة اتزان حرارى مع B فان الجسم A,C فى حالة اتزان حرارى عند التلامس.

4 1-2 Thermometers and temperature scale 1- Thermocouple ترمومتر الازدواج الحرارى Seebeck effect: The conversion of temperature differences directly into electricitytemperatureelectricity وجود فرق فى درجات الحرارة بين نقطتى اتصال معدنين مختلفين يولد قوة دافعة كهربية وتعتمد قيمتها على فرق درجة الحرارة.

5 Continue: يتكون ترمومتر الازدواج الحرارى من : 1- معدن A 2- معدن اخر B 3- فولتميتر لقياس فرق الجهد مميزاته : 1- الأنواع التجارية رخيصة الثمن 2- يقيس مدى واسع من درجات الحرارة

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7 2- A bimetallic strip thermometer ترمومتر ثنائى المعدن  يتكون ترمومتر ثنائى المعدن من شريحتين من معدنين مختلفين مثبتتين مع بعضهما. عندما تزيد درجة حرارة هذه الشريحة المركبة يكون الزيادة فى الطول لأحدى الشريحتين أكبر من الأخرى مما يسبب تقوس الشريحة كما بالشكل.

8 Continue لقياس درجة حرارة جسم ما يتم توصيل أحد طرفى الشريحة ثنائية المعدن بمؤشر يتحرك على تدريج والطرف الأخر يكون ملامسا للجسم. لقياس درجة حرارة جسم ما يتم توصيل أحد طرفى الشريحة ثنائية المعدن بمؤشر يتحرك على تدريج والطرف الأخر يكون ملامسا للجسم. عند ارتفاع درجة حرارة الجسم فان الشريحة تنحنى بحيث تسبب دوران المؤشر على التدريج. عند ارتفاع درجة حرارة الجسم فان الشريحة تنحنى بحيث تسبب دوران المؤشر على التدريج.

9 3- Resistance thermometer  ρ resistivity at temperature t c  ρ 20 resistivity at temperature 20C  α temperature coefficient of resistivity.

10 Continue

11 4- Thermopile thermometers  Don’t require physical contact with the object they are measuring. which measures the amount of infrared radiation emitted by the eardrum to indicate the temperature of the eardrum.  هذا الترمومترلايحتاج الى ملامسة الجسم المراد قياس درجة حرارته. فهو يقيس التأثير الحرارى للاشعة تحت الحمراء المنبعثة من الجسم.

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13 Temperature scale  The Celsius scale: in this scale the freezing point of water is 0°C and the boiling point is 100°C. التدريج المئوى اتخذ نقطة تجمد الماء صفر درجة مئوية، ودرجة الغليان 100 درجة مئوية.

14 Continue:  The Fahrenheit scale : in this scale water freezes at 32°F and boils at 212°F. التدريج الفهرنهيتى اتخذ نقطة تجمد الماء 32 درجة فهرنهيت، ودرجة الغليان 212 درجة فهرنهيت.

15 Continue Conversion between degrees Celsius and degrees Fahrenheit is given by T F = ( 9/5 ) T C + 32 °F T F = ( 9/5 ) T C + 32 °F T C = 5/9 ( T F – 32 °F ) T C = 5/9 ( T F – 32 °F )

16 Example 1-1 (a) On a fine spring day you notice that the temperature is 75°F. What is the corresponding temperature on the Celsius scale? (a) On a fine spring day you notice that the temperature is 75°F. What is the corresponding temperature on the Celsius scale? (b) If the temperature on a brisk winter morning is -2°C, what is the corresponding Fahrenheit temperature? (b) If the temperature on a brisk winter morning is -2°C, what is the corresponding Fahrenheit temperature?

17 Example 1-2 What temperature is the same on both the Celsius and Fahrenheit scales? What temperature is the same on both the Celsius and Fahrenheit scales? Solution : Set T F = T C = t T F = ( 9/5 ) T C + 32 t = ( 9/5 ) t + 32 t = ( 9/5 ) t + 32 t = - 40 t = - 40

18 Continue The Kelvin scale : the Kelvin scale is also chosen to have the same degree size as the Celsius scale. The Kelvin scale : the Kelvin scale is also chosen to have the same degree size as the Celsius scale. Conversion between degrees Celsius and degrees Kelvin is given by Conversion between degrees Celsius and degrees Kelvin is given by T = T C T = T C

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20 1-3 Thermal expansion Linear expansion  L = L   T Where α the coefficient of linear expansion α the coefficient of linear expansion has the unit C -1 or K -1 Before heating L After heating L ∆L

21 Example 1-3 A surveyor uses a steel measuring tape that is exactly 50.0m long at a temperature of 20°C. What is its length on a hot summer day when the temperature is 35°C. Solution  L = L   T = (1.2 x k -1 ) (50m) (15K) = 9 mm = 9 mm L = L o +  L = m

22 Table ( 1-1 ) Coefficient of thermal expansion near 20 °C 0.5Quartz 1.51Ether3.3Pyrex glass 0.18Mercury11Window glass 0.21Water12Concrete 0.68Olive oil12Iron ( steel ) 0.95Gasoline17Copper 1.01Alcohol19Brass 1.18Carbon tetrachloride 24Aluminum 0.5Quartz29Lead  ( / °C ) Substance  ( / °C ) Substance

23 Volume expansion V = V  T Where β is the coefficient of volume expansion of the solid or liquid.  = 3  for solid Before heating After heating

24 Example 1-4 On a hot day in Las Vegas, an oil trucker loaded L of diesel fuel. He encountered cold weather on the way to Payson, Utah, where the temperature was 23K lower than in Las Vegas, and where he delivered his entire load. How many liters did he deliver? The coefficient of volume expansion for diesel fuel is 9.5 x /°C and the coefficient of linear expansion for his steel truck tank is 11 x / °C.

25 Example (1-5 ) A copper flask with a volume of 150 cm 3 is filled to the brim with olive oil. If the temperature of the system is increased from 6 °C to 31 °C, how much oil spills from the flask?

26 Solution  V oil =  V  T = (0.68 x k -1 ) ( 150 cm 3 ) (25k)  V oil =  V  T = (0.68 x k -1 ) ( 150 cm 3 ) (25k) = 2.6 cm 3 = 2.6 cm 3  V flask = 3  V  T = 3 (17 x k -1 ) ( 150 cm 3 ) (25k)  V flask = 3  V  T = 3 (17 x k -1 ) ( 150 cm 3 ) (25k) = 0.19 cm 3 = 0.19 cm 3  V oil -  V flask = 2.6 cm cm 3 = 2.4 cm 3  V oil -  V flask = 2.6 cm cm 3 = 2.4 cm 3

27 Thermal stress Thermal stress After cooling L ∆L Before cooling L Clamped rod (  L / L o ) thermal =   T E = ( F/A )/ (  L /L o ) (  L / L o ) tension = (  L / L o ) thermal   T = - F / ( A E ) F / A =  thermal  thermal = - E  T

28 Example ( 1- 6 ) An aluminum cylinder 10 cm long, with a cross-sectional area of 20 cm 2, is to be used as a spacer between two steel walls. At 17.2 °C it just slips in between the walls. When it warms to 22.3°C, calculate the stress in the cylinder and the total force it exerts on each wall, assuming that the walls are perfectly rigid and a constant distance apart.

29 1-4 Quantity of heat  The heat flow means that one substance loss heat and the other gains heat. So, when they in thermal equilibrium Heat lost = Heat gained

30 Units of heat  The Calorie ( Cal ) : which is defined as the amount of heat required to raise to the temperature of one gram of water from 14.5 °C to 15.5 °C. The kilocalorie ( kcal ) = 1000 cal.  British thermal unit ( Btu ) : is the quantity of heat required to raise the temperature of one pound (weight) of water 1 °F from 63 °F to 64 °F.  1 cal = J  1 Btu = 778 ft.lb = 252 cal = 1055 J

31 Specific heat  The quantity of heat Q  m T  Q = c m T  Where c is a quantity, different for different materials, called Specific heat of the material.

32 Example (1 – 7) During about with the flu an 80-kg man ran a fever of 39°C instead of the normal body temperature of 37°C. Assuming that the human body is mostly water, how much heat is required to raise this temperature by that amount? (c w = 4190 J/kg.k)

33 Table ( 1-2 ) Specific heats at atmospheric pressure

34 Example (1 – 8) An electronic circuit element made of 23 mg of silicon. The electric current through it adds energy at the rate of 7.4 mW = 7.4 x J/s. If no heat transfer out of the element, at what rate does its temperature increase. ( c s = 705 J /kg.k )

35 Example (1 – 9) A Copper slug whose mass m c is 75 gm is heated in laboratory oven to a temperature T of 312 °C. The slug is then dropped into a glass beaker containing a mass m w = 220 gm of water and the mass of the beaker is 225 gm. Heat capacity c b of the beaker is 0.2 cal/g.k. The initial temperature of the water and the beaker is 12 °C. Assuming that the slug, beaker, and water are an isolated system and the water does not vaporize, find the final temperature T f of the system at thermal equilibrium.

36 Solution In this case : 1- The slug loses energy Q c = m c c c ( T f – T c ) 2- The water gains energy Q w = m w c w ( T f – T ) 3- The beaker gains energy Q b = mb c b ( T f – T b ) Q Lose = Q gain Q s = Q w + Q b c c m c T c + c b m b T b + c w m w T w T f = c c m c + m b c b +c w m w

37 Latent heat A transition from one phase to another is called a phase change. The quantity of heat that is added or removed from the system is called the latent heat. تغير حالة المادة من صورة الى أخرى مثل تحول الماء الى ثلج أو العكس يسمى phase change. كمية الحرارة اللازمة لتحويل المادة من صورة الى أخرى تسمى الحرارة الكامنة.

38 Types of latent heat : 1- Latent heat of fusion ( L f ) : الحرارة الكامنة للانصهار أو للتجمد Q = m L f 2- Latent heat of vaporization ( L v ) : الحرارة الكامنة للتبخر أو للتكثف Q = m L v

39 Table ( 1-3 ) Heats of fusion and vaporization

40 Heating ( cooling )curve of water a b c d e

41 Study the heating curve: 1- From point a b, the state is solid (ice) and reach 0 ⁰ C ( melting point ), heat added Q = m C ice ( t a - 0 ). 1- لانتقال المادة من الحالة a ( ثلج ) عند درجة حرارة t a الى الحالة b ( ثلج ) عند درجة صفر مئوئ ( درجة الانصهار ) فان كمية الحرارة المضافة تساوى m C ice ( t a - 0 ).

42 Continue 2- From point b c, the solid begin to change to liquid with constant temperature 0 ⁰ C, at point c the state is liquid, heat added Q = m L f. 2- لانتقال المادة من الحالة b ( ثلج ) الى الحالة c ( ماء ) مع ثبوت درجة الحرارة عند صفر مئوى ( درجة الانصهار ) فان كمية الحرارة المضافة تساوى m L f

43 Continue 3- From point c d, the state is liquid and temperature rises to 100 ⁰ C ( the boiling point ), heat added Q = m C water ( ) 3- لانتقال المادة من الحالة c ( ماء ) عند درجة 0 مئوى الى الحالة d ( ماء ) عند درجة 100 مئوى ( درجة الغليان ) فان كمية الحرارة المضافة تساوى m C w (100-0)

44 Continue 4- From point d e, the liquid begin to change to vapor with constant temperature, at point e the state is vapor, heat added Q = m Lv. 4- لانتقال المادة من الحالة d ( ماء ) الى الحالة e ( بخار ) مع ثبوت درجة الحرارة عند 100 درجة مئوية ( درجة الغليان ) فان كمية الحرارة المضافة تساوى m L v

45 Continue 5- From point e to any point, the temperature of vapor rises to t by adding a quantity of heat Q= m c v (t -100). 5- للانتقال من الحالة e ( بخار ) عند 100 درجة مئوية الى بخار عند درجة حرارة t أعلى من 100 درجة مئوية تكون كمية الحرارة المضافة تساوى mc steam (t – 100)

46 Heating ( cooling )curve of the ordinary materials

47 Example ( ) How much heat must be absorbed by ice of mass m = 720 gm at -10 ⁰ C to take it to liquid state at 15 ⁰ C? How much heat must be absorbed by ice of mass m = 720 gm at -10 ⁰ C to take it to liquid state at 15 ⁰ C? Q1Q1 Q2Q2 Q3Q3

48 Solution يمكن تقسيم كمية الحرارة المضافة Q الى ثلاث كميات Q 1, Q 2, Q 3 Q 1 -1 لرفع درجة حرارة الثلج من درجة -10C الى درجة 0.0C 15 Q 1 = m ice C ice (0 + 10) Q 2 -2 لتحويل الثلج الى ماء عند درجة 0.0C Q 2 = m ice L f Q 3 -3 لرفع درجة حرارة الماء من درجة 0C الى درجة 15C Q 3 = m w C w (15 - 0) Q = Q 1 + Q 2 +Q 3  300 kJ

49 Example ( ) A 0.5-kg block of metal with an initial temperature of 54.5C is dropped into a container holding 1.1kg of water at 20.0C. If the final temperature of the block-water system is 21.4C, what is the specific heat of the metal? Assume the container can be ignored, and that no heat is exchanged with the surroundings.

50 Solution m w = 1.1 kg, T w = 20C, m b = 0.5kg, T b = 54.5C, T = 21.4C Q gain = Q lose Q gain = m w C w ( T – T w ) Q lose = m b C b ( T – T b ) C b = 390 J/(kg.k)


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