3.4 Thermal Energy Change
3.4. Explain the relationship among temperature change
in a substance, the amount of heat transferred, the
amount (mass) of the substance, and the specific heat of
the substance. Page 474-478 in Text. Section 16.1
Learning Objectives:
What is Specific Heat? Heat Capacity?
Tmeperature?
When is energy converted from one form to another?
Vocabulary – During this unit you will learn and understand the definitions of the following terms. Some are review. Also use this
as a word bank for the notes and exercise.
Temperature
Substance
Even
Heat/ Thermal
Energy
Specific Heat
Capacity
Heat Capacity
Molecular Velocity
mass
Kinetic Energy
times
More/Less
Increase/decrease
PUTTING #’S TO HEAT TRANSFER- Three things effect how heat is exchanged. Take a look at the following
situations. Identify what makes the temperature change for the containers different.
The two things for this situation that are the same are:
A -The subtance or material being heated (Water in both)
B -The quantity of heat (One Bunsen burner for both)
The variable that is different is: mass
Containers with different quantities of the same material are supplied with
equal heat. Will the temps change equally? (Yes or No) Why? Different
masses heat differently.
This shows that varying MASS changes how the temperature will change.
More mass heats up (LESS/SLOWER) with the same quantity of heat added.
Less mass heats up (MORE/FASTER) with the same quantity of heat added.
Heat is absorbed like a sponge soaking up water. A larger sponge can accept more water with less change.
The larger mass of water requires (more or less) energy to change its temperature.
EXTREME CASE- Imagine trying to heat the amount of water in the ocean with a candle. Would the
temperature change be large enough to measure? The energy would increase but not enough to notice.
The two things for this situation that are the same are:
A -Mass
B - Heat
The variable that is different is: Material or Substance
Containers with the same masses of different substances are supplied
with equal heat. Will the temps change equally? (Yes or No) Why?
Different materials heat differently.
This shows that varying the substance changes how the temperature
(and therefore the thermal energy) will change.
The property of how a substance heats up is called the SPECIFIC HEAT CAPACITY. It is “specific” to the
substance. Specific heat is a measure of how a substance resists being heated. A sort of heat inertia.
Each substance has a unique specific heat.
The two things for this situation that are the same are:
A –Mass
B – Material
The variable that is different is: Quantity of Heat shown by the
number of flames under the beakers
Containers with the same masses of the same substance are heated
with unequal quantities of thermal energy. Will the temp change be
equal? (Yes or No) Why? The amount of heat is different.
This shows that varying the Energy going into the Substance changes
how the temperature (and therefore the thermal energy) will change.
More (heat) Thermal Energy creates a (larger or smaller) temperature change.
Less (heat) Thermal Energy creates a (larger or smaller) temperature change.
Specific Heat is C
SPECIFIC HEAT CAPACITY – Resistance to Temperature Change for a
substance
A high specific heat means the substance will not change temperature easily.
A low specific heat means the substance will change temperature easily.
Which will change temperature the most easily? Lead (Lowest Specific Heat)
Which will change temperature the least easily? Water (Highest Specific Heat)
Water ,which makes up a great deal of the surface of the Earth has a very high
specific heat .
HEAT CAPACITY = (m x C)
HEAT CAPACITY – Resistance to Temperature Change for a known mass of substance
Heat Capacity is a measure of how easily an object changes temperature. Heat Capacity is the mass times
Specific heat:
Heat Capacity = m x C
Heat capacity can be used to compare how easily two objects will change temperature.
EXAMPLE: Given equal quantities of thermal energy, which object will heat up more?
Heat Capacity for 2.5 kg of Water = (2.5 kg)x(4200 J/kgC) = 10,500 J/C
Heat Capacity for 1.5 kg of Lead = (1.5 kg)x(128 J/kgC) = 192 J/C
2.5 kg of
Water
1.5 kg of
Lead
The water will require more than 50 times the energy to raise its temperature the same amount.
The Q Equation - This equation connects, Heat Energy (Q), mass (m) ,
Specific Heat (C) and the change in Temperature (ΔT)
Q = m C ΔT
Q = Heat Energy ( J), m = mass ( kg or g),
C = Specific Heat ( J/kg C or J / g C), ΔT = Temperature Change ( degrees C)
Hewitt’s Webcasts:
HEAT:
http://conceptualacademy.org/course/conceptual-physics/152-heat
SPECIFIC HEAT CAPACITY:
http://conceptualacademy.org/course/conceptual-physics/153-specific-heat-capacity
LAST NAME __________________ FIRST NAME _____________________ DATE_____PERIOD____
3.4 SPECIFIC HEAT AND HEAT CAPACITY
Q is Thermal Energy is Energy so it has the units of Joules.
Specific Heat of a substance has the units of J/kg C or J/g C.
Water has a very high Specific Heat ( 4,200 J/kg C) this means it will require
( a lot or not much ) Energy to change its temperature.
Specific Heat has the units of J/kg C or J/g C.
Substances with ( high or low ) specific heats will heat up easily while
substance with ( high or low ) specific heats will be more difficult to heat up.
A substance with a lower specific heat will heat up (MORE or LESS) than a substance with a high specific
heat.
HEAT CAPACITY QUESTIONS: Heat Capacity = m x C
Which will require more energy to change its temperature, 1 kg of water or 10 kg of Copper?
Heat Capacity for 1 kg of Water = (1 kg)x(4200 J/kgC) = 4,200 J/C
Heat Capacity for 10 kg of Copper = (10 kg)x(385J/kgC) = 3850 J/C
The water has a higher Heat Capacity so it will resist temperature change more. It will require more energy to change
its temperature.
Determine which sample of the pairs will require more energy to change its temperature?
Sample A = 1 kg of Aluminum
Sample B = 3 kg of Copper
Sample A = 9 kg of Steel
Sample B = 1 kg of Water
Sample A = 2 kg of ice
Sample B = 10 kg of lead
(1kg)(900 J/ kg C) = 900
(3kg)(385 J/ kg C) = 1,155
The copper sample has a higher
value so it till require more energy.
(9kg)(452 J/ kg C) = 4,086
(1kg)(4200 J/ kg C) = 4,200
The water sample has a higher value
so it till require more energy.
(2kg)(2000 J/ kg C) = 4,000
(10kg)(128 J/ kg C) = 1,280
The ice sample has a higher value so
it till require more energy.
Sample A = 10 kg of wood
Sample B = 20 kg of copper
Sample A = 3 kg of Lead
Sample B = 1 kg of Steel
Sample A = 5 kg of Ice
Sample B = 3 kg of water
(10kg)(1700 J/ kg C) = 17,000
(20 kg)(385 J/ kg C) = 7,700
The wood sample has a higher value
so it till require more energy.
(3kg)(128 J/ kg C) = 384
(1kg)(452 J/ kg C) = 452
The steel sample has a higher value
so it till require more energy.
(5kg)(2,000 J/ kg C) = 10,000
(3kg)(4,200 J/ kg C) = 12,600
The water sample has a higher value
so it till require more energy.
Sample A = 3 kg of copper
Sample B = 1 kg of Ice
Sample A = 8 kg of Aluminum
Sample B = 2 kg of Water
Sample A = 20 kg of Copper
Sample B = 5 kg of Water
(3kg)(385J/ kg C) = 1,155
(1kg)(2,000 J/ kg C) = 2,000
The ice sample has a higher value so
it till require more energy.
(8kg)(900 J/ kg C) = 7,200
(2kg)(4,200 J/ kg C) = 8,400
The water sample has a higher value
so it till require more energy.
(20kg)(385 J/ kg C) = 7,700
(5 kg)(4,200 J/ kg C) = 21,000
The water sample has a higher value
so it till require more energy.
Which will be easier to keep warm, a small house or a large house? A small house will have less mass to
keep heated.
Do you think a house twice as large requires twice as much energy to keep it warm?
Q=mC(ΔT)
EXAMPLE: How much energy (Q) is required to heat 15 kg of water from 20 degrees to 40 Celsius?
List what you know Calculate
(with units)
m = 15 kg
C= 4200 J/(kg C)
ΔT = 20 C
State the Unknown
Q=mC(ΔT) = (15 kg)( 4200 J/(kg C))(20 C) = 1,260,000 J
Q=?
State the Equation
Q=mC(ΔT)
Solution 1,260,000 J
1. How much energy must be added to 6 kg of wood to increase its temperature from 30 to 40 C?
(102,000 J)
List what you know
Calculate
(with units)
m = 6 kg
C= 1700 J/(kg C)
ΔT = 10 C
Q=mC(ΔT) = (6 kg)( 1,700 J/(kg C))(10 C) = 102,000 J
Unknown
Q=?
State the Equation
Q=mC(ΔT)
2. How much energy must be removed from 10 kg of water to cool its temperature from 40 to 30 C?
(420,000 J)
List what you know
Calculate
(with units)
m = 10 kg
C= 4,200 J/(kg C)
ΔT = -10 C
Q=mC(ΔT) = (10 kg)( 4,200 J/(kg C))(-10 C) = -420,000 J
Unknown
Q=?
State the Equation
Q=mC(ΔT)
3. How much energy must be added to 150 g of water to heat it from 30 to 40 C? (6,300 J)
List what you know
Calculate
(with units)
m = 150 g
C= 4.2 J/(g C)
ΔT = 10 C
Unknown
Q=?
State the Equation
Q=mC(ΔT)
Q=mC(ΔT) = (150 g)( 4.2 J/(g C))(10 C) = 6,300 J
4. How much energy must be added to 200 g of copper to heat it from 25 to 40 C? (1,155 J)
List what you know
Calculate
(with units)
m = 200 g
C= .385 J/(g C)
ΔT = 15 C
Unknown
Q=?
State the Equation
Q=mC(ΔT)
Q=mC(ΔT) = (200 g)( .385 J/(g C))(10 C) = 1,155 J
CALCULTING TEMPERATURE CHANGE
EXAMPLE: How much will the temperature of 15 kg of water change if 1,260,00 J (Q) of heat is added?
List what you know
(with units)
m = 15 kg
C= 4200 J/(kg C)
Q = 1,260,000 J
State the Unknown
ΔT = ?
State the Equation
Q=mC(ΔT)
Δ𝑇 =
𝑄
𝑚𝐶
1,260,000 𝐽
15 𝑘𝑔)(4,200)
=(
= 20 𝐶
Solution 20 C
1. How much will the temperature of 1000 g of water change if 5,000 J (Q) of heat is added? (1.2 C)
List what you know
(with units)
State the Unknown
m = 1000 g
C= 4.2 J/(g C)
𝑄
5,000 𝐽
1000 𝑘𝑔)(4.2)
Q = 5,000 J
Δ𝑇 =
State the Unknown
ΔT = ?
State the Equation
Solution Round this answer to 1.2 C.
𝑚𝐶
=(
= 1.19 𝐶
Q=mC(ΔT)
If the water started at 20.0 C, what would be the final temperature? 20 + 1.2 = 21.2 (21.2 C)
2. How much will the temperature of 10 kg of copper change if 385,000 J (Q) of heat is added? (100 C)
List what you know
(with units)
State the Unknown
m = 10 kg
C= 385 J/(kg C)
Q = 385,000 J
Δ𝑇 =
State the Unknown
ΔT = ?
State the Equation
Solution
Q=mC(ΔT)
𝑄
𝑚𝐶
385,000 𝐽
10 𝑘𝑔)(385)
=(
= 100 𝐶
3. How much will the temperature of 200 kg of steel change if 9,040,000 J (Q) of heat is added? (100 C)
List what you know
(with units)
State the Unknown
m = 200 kg
C= 452 J/(kg C)
Q = 9,040,000 J
Δ𝑇 =
State the Unknown
ΔT = ?
State the Equation
Solution
Q=mC(ΔT)
𝑄
𝑚𝐶
9,040,000 𝐽
200 𝑘𝑔)(452)
=(
= 100 𝐶