Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter Eleven: Heat  11.1 Heat  11.2 Heat Transfer.

Similar presentations


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

1

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.

9

10 11.1 The heat equation

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

12 1.Looking for:  …amount of heat in joules 2.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 3.Relationships:  E = mC p (T 2 – T 1 ) Solving Problems 4.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 10 7 J


Download ppt "Chapter Eleven: Heat  11.1 Heat  11.2 Heat Transfer."

Similar presentations


Ads by Google