Transcript Standard EPS Shell Presentation

HEAT 11.1
Chapter Eleven: Heat
11.1 Heat
11.2 Heat Transfer
Chapter 11.1 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.
Investigation 11A
Temperature and Heat
Key Question:
How are temperature and heat related?
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.
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.
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.
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.
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.
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.
Solving Problems
 How much heat is needed to raise the
temperature of a 250-liter hot tub
from 20°C to 40°C?
Solving Problems
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 = mCp(T2 – T1)
4. Solution:
Sig.
E =fig./Sci.
(250L ×not.
1kg/L)
× 4,184 J/kg°C
=
20,920,000
J = 2.1(40°C
x 10-7 20°C)
J
20,920,000 J