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Thermal Equilibrium

  • Temperature can be interpreted as the average kinetic energy of the molecules of a material
  • High temperature corresponds to high kinetic energy
  • A thermometer is a device to measure the temperature of an object

  • When you put a hot body ‘A’ in contact with a cold body ‘B’, interactions between these two bodies will occur eliminating the temperature difference between them
  • At the point where there is no more interaction, the system of the two bodies is said to be in a steady state which is called thermal equilibrium
  • At thermal equilibrium, ‘A’ and ‘B’ have the same temperature
  • Heat flow or heat transfer is the energy transfer that happens only due to a temperature difference
  • The Zeroth Law of Thermodynamics states that if ‘C’ is in thermal equilibrium with ‘A’ and ‘B’, then ‘A’ is also in thermal equilibrium with ‘B’


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Temperature Scales

1. Celsius Temperature Scale
  • We set zero to the freezing temperature of pure water
  • We set 100 to the boiling point of pure water
  • Divide the distance between these two points into 100 equal degrees
2. Fahrenheit Temperature Scale
  • The freezing temperature of water is 32 ºF
  • The boiling temperature of water is 212 ºF
  • The conversion relation between temperature in degrees of Celsius and degrees of Fahrenheit is:
TF=95TC+32°T_F=\frac{9}{5}T_C+32\degree
3. Kelvin Temperature Scale
  • The units are the same size as those in the Celsius scale
  • The zero of temperature at Kelvin scale is equal to 273.15°C-273.15\degree C. This point is called absolute zero
  • The relation between temperature in Kelvin scale and Celsius scale is:
TK=TC+273.15°T_K=T_C+273.15\degree
  • Kelvin is the standard unit of temperature
  • Since the size of units are the same in Kelvin and Celsius temperature scales:
ΔTK=ΔTC\Delta T_K=\Delta T_C



The temperature difference between the inside and outside of a home on a cold winter day is 50oF. Express the difference on

a) Celsius Scale
b) Kelvin Scale

a) TF=95TC+32°FT_F=\frac{9}{5}T_C+32\degree F

TC=59(TF32°F)T_C=\frac{5}{9}\left(T_F-32\degree F\right)

TC=59(5032)T_C=\frac{5}{9}\left(50-32\right)

TC=59(18)2T_C=\frac{5}{\cancel{9}}\left(\cancel{18}\right)^{2}

TC=10°CT_C=10\degree C

b) TC=T273.15T_C=T-273.15

T=TC+273.15T=T_C+273.15

T=10+273.15T=10+273.15

T=283.15°KT=283.15\degree K
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Peter was not feeling well. He checked his body temperature using a thermometer, it was 40oC. What is this temperature on the Fahrenheit scale? Do you think Peter is seriously ill!

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