Ch. 5 - Energy II. Thermal Energy (p.134-137, 141-144)  Temperature  Thermal Energy  Heat Transfer.

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Presentation transcript:

Ch. 5 - Energy II. Thermal Energy (p , )  Temperature  Thermal Energy  Heat Transfer

A. Temperature Temperature  measure of the average KE of the particles in a sample of matter

B. Thermal Energy Thermal Energy  the total energy of the particles in a material  KE - movement of particles  PE - forces within or between particles due to position  depends on temperature, mass, and type of substance

B. Thermal Energy Which beaker of water has more thermal energy?  B - same temperature, more mass 200 mL 80ºC A 400 mL 80ºC B

C. Heat Transfer Heat  thermal energy that flows from a warmer material to a cooler material Like work, heat is...  measured in joules (J)  a transfer of energy

C. Heat Transfer Why does A feel hot and B feel cold? 80ºC A 10ºC B  Heat flows from A to your hand = hot.  Heat flows from your hand to B = cold.

C. Heat Transfer Specific Heat (C p )  amount of energy required to raise the temp. of 1 kg of material by 1 degree Kelvin  units: J/(kg·K) or J/(kg·°C)

C. Heat Transfer Which sample will take longer to heat to 100°C? 50 g Al50 g Cu Al - It has a higher specific heat. Al will also take longer to cool down.

C. Heat Transfer Q = m   T  C p Q:heat (J) m:mass (kg)  T:change in temperature (K or °C) C p :specific heat (J/kg·K)  T = T f - T i – Q = heat loss + Q = heat gain

C. Heat Transfer Calorimeter  device used to measure changes in thermal energy Coffee cup Calorimeter  in an insulated system, heat gained = heat lost

C. Heat Transfer A 32-g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon? GIVEN: m = 32 g T i = 60°C T f = 20°C Q = ? C p = 235 J/kg·K WORK: Q = m·  T·C p m = 32 g = kg  T = 20°C - 60°C = – 40°C Q = (0.032kg)(-40°C)(235J/kg·K) Q = – 301 J

C. Heat Transfer How much heat is required to warm 230 g of water from 12°C to 90°C? GIVEN: m = 230 g T i = 12°C T f = 90°C Q = ? C p = 4184 J/kg·K WORK: Q = m·  T·C p m = 230 g = 0.23 kg  T = 90°C - 12°C = 78°C Q = (0.23kg)(78°C)(4184 J/kg·K) Q = 75,061 J