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 equilibrium with object B, then objects A and C will be in thermal equilibrium if brought into thermal contact.

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.

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

Continue: يتكون ترمومتر الازدواج الحرارى من: مميزاته: 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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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.

Continue 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.

3- Resistance thermometer ρ resistivity at temperature tc ρ20 resistivity at temperature 20C α temperature coefficient of resistivity.

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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. www.youtube.com/watch?v=8nAypo1mXX4

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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 100درجة مئوية.

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 212درجة فهرنهيت.

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

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? (b) If the temperature on a brisk winter morning is -2°C, what is the corresponding Fahrenheit temperature?

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

Continue 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 T = TC + 273.15

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

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 10-5 k-1) (50m) (15K) = 9 mm L = Lo + L = 50.009 m

Table ( 1-1 ) Coefficient of thermal expansion near 20 °C Substance  ( 10-6 / °C ) Substance  ( 10-3/ °C ) Lead 29 Quartz 0.5 Aluminum 24 Carbon tetrachloride 1.18 Brass 19 Alcohol 1.01 Copper 17 Gasoline 0.95 Iron ( steel ) 12 Olive oil 0.68 Concrete 12 Water 0.21 Window glass 11 Mercury 0.18 Pyrex glass 3.3 Ether 1.51 Quartz 0.5

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

Example 1-4 On a hot day in Las Vegas, an oil trucker loaded 37000 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 10-4/°C and the coefficient of linear expansion for his steel truck tank is 11 x 10-6 / °C.

Example (1-5 ) A copper flask with a volume of 150 cm3 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?

Solution Voil =  V T = (0.68 x 10-3 k-1) ( 150 cm3) (25k) = 2.6 cm3 Vflask = 3  V T = 3 (17 x 10-6 k-1) ( 150 cm3) (25k) = 0.19 cm3 Voil - Vflask = 2.6 cm3 - 0.19 cm3 = 2.4 cm3

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

Example ( 1- 6 ) An aluminum cylinder 10 cm long, with a cross-sectional area of 20 cm2, 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.

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

1- 4 -1 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 = 4.186 J 1 Btu = 778 ft.lb = 252 cal = 1055 J

1- 4 -2 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.

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? (cw = 4190 J/kg.k)

Table ( 1-2 ) Specific heats at atmospheric pressure

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 10-3 J/s. If no heat transfer out of the element, at what rate does its temperature increase. ( cs = 705 J /kg.k )

Example (1 – 9) A Copper slug whose mass mc 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 mw = 220 gm of water and the mass of the beaker is 225 gm. Heat capacity cb 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 Tf of the system at thermal equilibrium.

Solution In this case : 1- The slug loses energy Qc = mc cc ( Tf – Tc ) 2- The water gains energy Qw = mw cw ( Tf – T ) 3- The beaker gains energy Qb = mb cb ( Tf – Tb ) QLose = Qgain Qs = Qw + Qb cc mc Tc + cb mb Tb + cw mw Tw Tf = cc mc + mb cb +cw mw

1-4 -2 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.

Types of latent heat : 1- Latent heat of fusion ( Lf ) : الحرارة الكامنة للانصهار أو للتجمد Q = m Lf 2- Latent heat of vaporization ( Lv ) : الحرارة الكامنة للتبخر أو للتكثف Q = m Lv

Table ( 1-3 ) Heats of fusion and vaporization

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

Study the heating curve: 1- From point a b, the state is solid (ice) and reach 0⁰C ( melting point ), heat added Q = m Cice ( ta - 0 ). 1- لانتقال المادة من الحالة a (ثلج) عند درجة حرارة ta الى الحالة b (ثلج) عند درجة صفر مئوئ (درجة الانصهار) فان كمية الحرارة المضافة تساوى m Cice ( ta - 0 ).

Continue 2- From point b c, the solid begin to change to liquid with constant temperature 0⁰C, at point c the state is liquid, heat added Q = m Lf . 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 Lf

Continue 3- From point c d, the state is liquid and temperature rises to 100⁰C ( the boiling point ), heat added Q = m Cwater ( 100 - 0 ) 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 Cw (100-0)

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 100درجة مئوية (درجة الغليان) فان كمية الحرارة المضافة تساوى m Lv

Continue 5- From point e to any point, the temperature of vapor rises to t by adding a quantity of heat Q= m cv(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 mcsteam(t – 100)

Heating ( cooling )curve of the ordinary materials

Example ( 1- 10 ) How much heat must be absorbed by ice of mass m = 720 gm at -10⁰C to take it to liquid state at 15⁰C? 15 -10 Q1 Q2 Q3

Solution Q1 -1 لرفع درجة حرارة الثلج من درجة -10C الى درجة 0.0C 15 يمكن تقسيم كمية الحرارة المضافة Qالى ثلاث كميات Q1, Q2, Q3 Q1 -1 لرفع درجة حرارة الثلج من درجة -10C الى درجة 0.0C 15 Q1 = mice Cice (0 + 10) Q2 -2 لتحويل الثلج الى ماء عند درجة 0.0C Q2 = mice Lf Q3 -3 لرفع درجة حرارة الماء من درجة 0C الى درجة 15C Q3 = mw Cw (15 - 0) Q = Q1 + Q2 +Q3  300 kJ

Example ( 1 -11 ) 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.

Solution mw = 1.1 kg, Tw = 20C, mb = 0.5kg, Tb = 54.5C, T = 21.4C Qgain = Qlose Qgain = mw Cw ( T – Tw ) Qlose = mb Cb ( T – Tb ) Cb = 390 J/(kg.k)