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Chapter Eleven: Heat 11.1 Heat 11.2 Heat Transfer.

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Presentation on theme: "Chapter Eleven: Heat 11.1 Heat 11.2 Heat Transfer."— Presentation transcript:

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2 Chapter Eleven: Heat 11.1 Heat 11.2 Heat Transfer

3 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.

4 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.

5 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.

6 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.

7 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.

8 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.

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10 11.1 The heat equation

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

12 Solving Problems Looking for: Given: Relationships: Solution:
…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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