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Chapter Eleven: Heat

Chapter Eleven: Heat. 11.1 Heat 11.2 Heat Transfer. 11.1 What is heat?. Heat is thermal energy that is moving. Heat flows any time there is a difference in temperature.

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Chapter Eleven: Heat

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  1. Chapter Eleven: Heat • 11.1 Heat • 11.2 Heat Transfer

  2. 11.1 What is heat? • Heat is thermal energy that is moving. • Heat flows any time there is a difference in temperature. • Because your hand has more thermal energy than chocolate, thermal energy flows from your hand to the chocolate and the chocolate begins to melt.

  3. 11.1 What is heat? • Heat and temperature are related, but are not the same thing. • The amount of thermal energy depends on the temperature but it also depends on the amount of matter you have.

  4. 11.1 Units of heat and thermal energy • The metric unit for measuring heat is the joule. • This is the same joule used to measure all forms of energy, not just heat.

  5. 11.1 Heat and thermal energy • Thermal energy is often measured in calories. • One calorie is the amount of energy it takes to raise the temperature of one milliliter of water by one degree Celsius.

  6. 11.1 Specific heat • The specific heat is a property of a substance that tells us how much heat is needed to raise the temperature of one kilogram of a material by one degree Celsius. Knowing the specific heat of a material tells you how quickly the temperature will change as it gains or loses energy.

  7. 11.1 Why is specific heat different for different materials? • Temperature measures the average kinetic energy per particle. • Energy that is divided between fewer particles means more energy per particle, and therefore more temperature change. • In general, materials made up of heavy atoms or molecules have low specific heat compared with materials made up of lighter ones.

  8. 11.1 The heat equation

  9. Solving Problems • How much heat is needed to raise the temperature of a 250-liter hot tub from 20°C to 40°C?

  10. Solving Problems • Looking for: • …amount of heat in joules • Given: • V = 250 L, 1 L of water = 1 kg • Temp changes from 20°C to 40°C • Table specific heat water = 4, 184 J/kg°C • Relationships: • E = mCp(T2 – T1) • Solution: • E = (250L × 1kg/L) × 4,184 J/kg°C (40°C - 20°C) = 20,920,000 J • Sig. fig./Sci. not. 20,920,000 J = 2.1 x 107 J

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