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Heat capacity and specific heat capacity

Heat capacity and specific heat capacity. Heat Capacity. The amount of heat required to raise the temperature of a body by 1 o C Energy (E) Heat Capacity (C) = --------------------------- temp. change ( )

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Heat capacity and specific heat capacity

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  1. Heat capacity and specific heat capacity

  2. Heat Capacity • The amount of heat required to raise the temperature of a body by 1 oC Energy (E) Heat Capacity (C) = --------------------------- temp. change () Unit : J/ oC or J oC-1

  3. Exampleinitial joulemeter reading = 30000 Jfinal joulemeter reading = 40000 Jinitial temperature = 25oC , final temperature = 35oCincrease in temperature = 10oC40000 - 30000 Heat Capacity C = ----------------------- 10 10000 = --------- = 1000 J/oC 10 Assumption : no heat loss to surroundings

  4. Specific Heat Capacity Heat energy required to heat a substance depends on • the mass of the substance • the temperature rise • the nature of the substance

  5. Comparison among different s.h.c. higher s.h.c. => difficult to raise its own temperature

  6. Definition of Specific Heat Capacity • The amount of heat required to raise the temperature of a body by 1 oC per kg Specific Heat Energy (E) Capacity (s.h.c ) = --------------------------------- temp. change () x mass (m) Unit : J/ oC /kg or J oC-1 kg-1

  7. QuestionHow many energy is required to raise 2 kg of water for 10oC?(s.h.c. of water is 4200 J/kg/oC )heat gained by water = m • c •  = 2 • 4200 • 10 = 84000 JAssumption : no heat loss to surroundings

  8. QuestionA 100 W heater heats up a metal block of mass 2 kg for 5 minutes, the temperature of the metal block increases from 25oC to 35oC. What is the s.h.c. of the metal block?heat supplied by heater = P • theat gained by metal block = m •c • By conservation of energyheat lost by heater = heat gained by metal block100 • 5 • 60 = 2 • c • (35 - 25) c = 1500 J/kg/oC Assumption : no heat loss to surroundings

  9. Which one has a higher temperature? Body 1 or Body 2

  10. Which one is the heat gain ? Which one is the heat lost ?

  11. Do both bodies have the same temperature?

  12. Calculation • Assumption • no heat loss to surroundings, thermometer stirrer and beaker By conservation of energy Heat gained = Heat Loss m2c2( - 2) = m1c1(1 - ) where  is the final temperature

  13. Question0.2 kg of water at 59oC is added to 0.4 kg of water at 20oC in a polystyrene cup. What is the final temperature of the mixture?(s.h.c. of water is 4200 J/kg/oC )heat gained by 20oC water = m • c •  = 0.4 • 4200 • (T-20) heat lost by 59oC water = m • c •  = 0.2 • 4200 • (59 -T) heat gained = heat lost Therefore, T = 33 oC Why is the polystyrene cup used? To reduce the heat loss to the surrounding

  14. Practical importance of high s.h.c. of water • s.h.c. of water is 4200 J oC-1 kg-1 , it isrelatively high • Damping of temperature change by water • for a fixed amount of energy absorbed, water has a much smaller temperature rise due to high s.h.c. • Damping of temperature change in living organisms • living organism, with high percentage of water, responds slowly to temperature change of the surroundings

  15. Practical importance of high s.h.c. of water • Damping of climate change • the temperature change of sea changes much more slowly than that of the land so that the sea are relatively • cooler summer & milder winter • cooler daytime & milder night

  16. Practical importance of high s.h.c. of water • Use of water as a coolant • Water can be used as • radiator coolant in motor car • coolant in nuclear reactor • Water is circulated around the pipes of the devices absorbing heat

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