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Phys 2101 Gabriela González

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Zeroth law of thermodynamics: If bodies A and B are each in thermal equilibrium with a third body T, then they are in thermal equilibrium with each other.

Or: every body in thermal equilibrium has a well defined “temperature”. When two bodies have the same temperature, they are in thermal equilibrium.

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•  On small scale, temperature is a measure of the vibration of the atoms in the material (solid, gas or fluid).

•  SI unit: kelvin (K)   There is an absolute minimum temperature (when all the atoms are still). That temperature is zero

kelvin.   There is a special temperature at which ice, water and vapor coexist: “triple point of water”. This

temperature is defined as 273.16 K. •  Other common scales:

  Celsius (oC): Tc=T-273.15. ice-water temperature =0oC, boiling water =100oC.   Fahrenheit (oF): TF=(9/5)TC+32. (0oF=brine temperature; 100oF~human body temperature)

a)  Normal Body temperature is 98.6 oF. What is it in kelvins and oC?

b)  Comfort level is usually accepted to be 68oF. What is it in kelvins and oC?

Fahrenheit, Daniel Gabriel (1686-1736)

Lord Kelvin (1824-1907)

Anders Celsius 1701-1744

Tc=(98.6-32)*5/9=37 oC Tk=37+273.15=310.15 K

Tc=(68-32)*5/9=20 oC Tk=20+273.15=293.15 K

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When temperature changes, solid bodies change their length, and all bodies change their volume according to:

ΔL = L α ΔT with α = coefficient of linear expansion

ΔV = V β ΔT with β = coefficient of volume expansion

•  If a body has a length L0 at temperature T0, what is its length at a temperature T?

L(t) = L0+ΔL = L0+L0αΔT = L0 (1+αΔT)

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A highway is made of concrete slabs that are 15m long at 20oC.

(a)  If the temperature range at the location of the highway is from -20oC to +40oC, what size expansion gap should be left (measured at 20oC) to prevent buckling of the highway? Concrete has a coefficient of linear expansion α=12 x 10-6 K-1

(b) How large are the gaps at -20oC? (c) What is the temperature range in

degrees Fahrenheit?

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A highway is made of concrete slabs that are 15m long at 20oC. (a)  If the temperature range at the location of the highway is from -20oC to +40oC, what size expansion gap

should be left (measured at 20oC) to prevent buckling of the highway? Concrete has a coefficient of linear expansion α=12 x 10-6 K-1

(b)  How large are the gaps at -20oC? (c)  What is the temperature range in degrees Fahrenheit?

At -20oC, ΔL=15m x 12 x 10-6 K-1 (-20oC-20oC)) = -7.2 mm At +40oC, ΔL=15m x 12 x 10-6 K-1 (40oC-20oC)) = +3.6 mm

(a)  The gaps need to be 3.6 mm measured at 20oC.

(b)  The gaps will be 10.8mm at -20oC.

(c)  A difference of 60oC is the same as (9/5)*60=108oF.

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The coefficient α is usually a (weak) function of temperature, and is usually positive (materials expand when temperature increases). If α is positive, the volume increases and the density decreases with temperature.

Is ice more or less dense than water?

Water at low temperatures is an exception! Water at 4oC is more dense than at the lower, freezing temperature of 0oC. This is why ice floats on water, and why there is ice on top of water in “frozen” lakes.

Water density vs. temperature

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Temperature is a property of systems in thermal equilibrium.

Heat is energy transferred between systems at different temperatures. (Temperature is to heat what velocity is to kinetic energy)

When talking about heat, we usually use the symbol Q. The SI unit for heat is the Joule. However, more common units are Btu (British thermal unit) and calories: 1 Btu = 1055 J 1 cal = 4.186 J (1 Cal= 1kcal = 1,000 cal) Air conditioners are 5000-12000 Btus; 10000 Btus ~ 10 million joules A “normal” diet has 1500-2000 Cal; 2000 Cal ~ 8 million joules

The heat capacity C of an object is the proportionality constant between heat absorbed and resulting change in temperature: Q = C ΔT = C (Tf-Ti) SI units for C are J/k, but common units are cal/oC.

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Heat capacity, like volume or mass, is a property that depends on the amount of material we are considering. A property that only depends on the substance (like density) is specific heat: c=C/m → Q = c m ΔT

The specific heat of water is c = 1 cal/(goC) = 1 Btu/(lboF) = 4190 J/kg/K

The specific heat of ice is different! cice=2220 J/kg/K

When a material undergoes a phase transformation (it is melting, or boiling), the temperature will not change, but heat is absorbed (or emitted) in the transformation. The energy per unit of mass is called the heat of transformation L: Q = L m

For water, the heat of vaporization is 2256 kJ/kg, and the heat of fusion is 333 kJ/kg.

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What is the amount of energy required to melt 100g of silver at room temperature? Silver melting temperature is 1235 K; specific heat is 236 (J/kgK); heat of fusion is 105 kJ/kg.

•  First, we need to get the silver to the melting temperature from room temperature, 293K: Q1= cAgmΔT = 236 (J/kgK) x 0.1 kg x (1235K-293K) =22,231 J

•  Then, we need to melt the silver: Q2 = Lm = 105 kJ/kg x 0.1kg = 10,500 J

Total energy required: Q= Q1+Q2 = 32,731 J = 31 Btu = 7.8 Cal

A--Granulating ladles. B--Wind furnace. C--Another wind furnace made of potter's clay, on a tripod. D--Crucible in which the silver is melted. E--Crucible used as ladle. F--Copper basin for drying the granules. G--An iron grate on which to heat the silver. H--The man who granulates. K--The man holding the broom. Treatise on the foremost mineralogy and metallurgy (1598)

http://oldsite.library.upenn.edu/etext/collections/smith/ercker/

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A person makes iced tea mixing 500g of hot tea (essentially water) with an equal mass of ice at -15oC. The tea’s initial temperature is 90oC, and after a while thermal equilibrium is reached.

a)  What is the mixture’s final temperature? b)  What are the final masses of ice and water in the

mixture?

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A person makes iced tea mixing 500g of hot tea (essentially water) with 50g of ice at -15oC. The tea’s initial temperature is 85oC, and after a while thermal equilibrium is reached. a)  What is the mixture’s final temperature? b)  What’s the final temperature if the initial mass of ice is also 500g?

(a) Let’s first assume all the ice melts, and the final temperature is some temperature T>0oC.

The heat absorbed by the melting ice is equal to the heat released by the cooling water. The ice first warms up to melting temperature, and needs: Q1 = miceciceΔT = 0.05 kg x 2.220 kJ/kg/K x 15K = 1.67 kJ Then the ice would need more heat to melt: Q2=0.05 kg x 333 kJ/kg = 16.7 kJ And then, the melted ice (now water) would need more heat to reach the final temperature T: Q3=0.05 kg x 4.190 J/kg/C (T-0oC) = 0.21 kJ/oC T

The ice released some heat while cooling down to temperature T (notice Q4<0!!) Q4=0.5 kg x 4.190 J/kg/oC (T-85oC) = 2.10 kJ/oC T – 178 kJ

Energy balance equation: 0 = Q1+Q2+Q3+Q4 = 1.67 kJ + 16.7 kJ + 0.21 kJ T/oC + 2.10 kJ T/oC -178 kJ = -160 kJ + 2.31 kJ T/C => T=69oC

(b) If the ice was 0.5 kg instead of 0.05 kg, the equation is: 0 = Q1+Q2+Q3+Q4 = 16.7 kJ + 167 kJ + 2.1 kJ T/oC + 2.10 kJ T/oC -178 kJ = 57 kJ + 4.2 kJ T/C => T=-13.6oC The final state is not water! The heat that can be released by the 500g of water cooling down to 0C is 178 kJ; the 500g of ice needs 16.7 kJ to warm up to 0C: there’s enough for that, leaving 161 kJ to melt the ice. But to melt all of 500g of ice, we need 167 kJ: there’s not enough heat in the water! The final state will be a mixture of ice and water. How much ice was melted? We have 161 kJ to melt ice, so m=161kJ/(333kJ/kg)=483 g will melt, there will be 17g of ice left floating in the (very cold) tea.