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

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

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Presentation transcript:

Ch. 5 - Energy II. Thermal Energy (p.134-137, 141-144) Temperature 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 A B Which beaker of water has more thermal energy? B - same temperature, more mass 200 mL 80ºC A 400 mL 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 A B Why does A feel hot and B feel cold? Heat flows from A to your hand = hot. Heat flows from your hand to B = cold. 80ºC A 10ºC B

C. Heat Transfer Specific Heat (Cp) 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 Al 50 g Cu Al - It has a higher specific heat. Al will also take longer to cool down.

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

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

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

C. Heat Transfer GIVEN: WORK: m = 230 g Q = m·T·Cp Ti = 12°C How much heat is required to warm 230 g of water from 12°C to 90°C? GIVEN: m = 230 g Ti = 12°C Tf = 90°C Q = ? Cp= 4184 J/kg·K WORK: Q = m·T·Cp 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