Chapter 9.1 Learning Goals Describe the relationship between heat, temperature, and thermal energy. Identify and use different units to measure heat. Explain how the specific heat of different materials can be used to describe changes in temperature and energy.

Temperature and Heat Key Question: Investigation 9A How are temperature and heat related?

9.1 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.

9.1 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.

9.1 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.

9.1 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.

9.1 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.

9.1 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.

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

Looking for: Given: Relationships: Solution: 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

Specific Heat of a Metal Investigation 9C Specific Heat of a Metal Key Question: How can you use specific heat to identify an unknown metal sample??