Lecture 22: Temperature and Thermal Expansion Internal Energy Temperature 0th Law of Thermodynamics Thermal Expansion/Contraction 1

Internal Energy and Temperature All objects have “internal energy” (measured in Joules): random motion of molecules kinetic energy collisions of molecules give rise to pressure Amount of internal energy depends on: temperature related to average kinetic energy per molecule how many molecules mass “specific heat” related to how many different ways a molecule can move translation rotation vibration the more ways it can move, the higher the specific heat

0th Law of Thermodynamics If two objects are in thermal equilibrium with a third, then the two are in thermal equilibrium with each other. If they are in thermal equilibrium, they are at the same temperature. Perhaps do hot water, and cold, comment on what happens after time.

Temperature Fahrenheit Celsius Kelvin 100 373.15 212 Water boils 32 Fahrenheit 100 Celsius 273.15 373.15 Kelvin Water boils Water freezes NOTE: K = 0 is “absolute zero”, meaning (almost) zero KE/molecule

Thermal Expansion When temperature rises: molecules have more kinetic energy (moving faster) consequently, things tend to expand Amount of expansion depends on… change in temperature original length coefficient of thermal expansion L =  L0 T (linear expansion) V =  V0 T (volume expansion) Temp: T Temp: T+T L0 L

Exception: Water Water is very unusual in that it has a maximum density at 4 ºC. That is why ice floats (and how we exist)!

Summary Temperature Thermal Expansion Measure of average Kinetic Energy of molecules Thermal Expansion L =  L0 T (linear expansion) V =  L0 T (volume expansion)

Example A brass rivet with diameter 1.0000 cm and a steel plate that has a hole with diameter 0.9995 cm are both at 20ºC. What temperature do they need to cooled or heated to for the rivet to fit in the hole? (brass = 19×10-6 ºC-1 and steel = 12×10-6 ºC-1) We will use the equation for thermal expansion/contraction: L =  L0 T

Example Ls + Ls = Lb + Lb Ls + s Ls Ts = Lb + b Lb Tb A brass rivet with diameter 1.0000 cm and a steel plate that has a hole with diameter 0.9995 cm are both at 20ºC. What temperature do they need to cooled or heated to for the rivet to fit in the hole? (brass = 19×10-6 ºC-1 and steel = 12×10-6 ºC-1) We want the two lengths to be equal: Ls + Ls = Lb + Lb Ls + s Ls Ts = Lb + b Lb Tb Ls + s Ls (Tf-Ti) = Lb + b Lb (Tf-Ti)

Ls + s Ls (Tf-Ti) = Lb + b Lb (Tf-Ti) Example A brass rivet with diameter 1.0000 cm and a steel plate that has a hole with diameter 0.9995 cm are both at 20ºC. What temperature do they need to cooled or heated to for the rivet to fit in the hole? (brass = 19×10-6 ºC-1 and steel = 12×10-6 ºC-1) Put in the known values and solve for Tf: Ls + s Ls (Tf-Ti) = Lb + b Lb (Tf-Ti) Tf = -51ºC