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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