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Section 7.2—Calorimetry & Heat Capacity
Why do some things get hot more quickly than others?
Temperature
Temperature – proportional to the average kinetic energy of the molecules
Energy due to motion(Related to how fast the molecules are moving)
As temperature increases
Molecules move faster
Heat & Enthalpy
Heat (q)– The flow of energy from higher temperature particles to lower temperature particles
Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”
Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume
Energy Units
The most common energy units are Joules (J) and calories (cal)
4.18 J 1.00 cal
1000 J
1000 cal
1 kJ
1 Cal (food calorie)
=
=
=
Energy Equivalents
These equivalents can be used in dimensional analysis to convert units
Heat Capacity
Specific Heat Capacity (C) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C
Cw (C for liquid water) =
1.00 cal/g°C or 4.18 J/g°C
Heat Capacity
High Heat Capacity Low Heat Capacity
Takes a large amount of energy to noticeably change temp
Small amount of energy can noticeably change temperature
Heats up slowly
Cools down slowly
Maintains temp better with small condition changes
Heats up quickly
Cools down quickly
Quickly readjusts to new conditions
A pool takes a long time to warm up and remains fairly warm over night.The air warms quickly on a sunny day, but cools quickly at night
A cast-iron pan stays hot for a long time after removing from oven.Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly
What things affect temperature change?
Heat Capacity of substanceThe higher the heat capacity, the slower the
temperature change
Mass of sampleThe larger the mass, the more molecules there are to
absorb energy, so the slower the temperature change
TCmH p
Energy added or removed
Mass of sample
Specific heat capacity of substance
Change in temperature
Positive & Negative T
Change in temperature (T) is always final temperature – initial temperature If temperature increases, T will be positive
A substance goes from 15°C to 25°C. 25°C - 15°C = 10°CThis is an increase of 10°C
If temperature decreases, T will be negativeA substance goes from 50°C to 35°C35°C – 50°C = -15°CThis is a decrease of 15°C
Positive & Negative H
Energy must be put in for temperature to increaseA “+” T will have a “+” H
Energy must be removed for temperature to decreaseA “-” T will have a “-” H
Let’s Practice #1
Example:How many joules must be
removed from 25 g of water at 75°C to drop the
temperature to 30°? Cw = 4.18 J/g°C
Let’s Practice #1
€
H = m ×Cw × ΔT
H = change in energym = massCw = heat capacity of water T = change in temperature (Tf - Ti)
H = - 4703J
CCg
JgH 0.750.3018.40.25
Example:How many joules must be
removed from 25.0 g of water at 75.0°C to drop the
temperature to 30.0°? Cp water = 4.18 J/g°C
Let’s Practice #2
If the specific heat capacity of aluminum is 0.900 J/g°C, how much heat must be added to a
30.0 g sample at 15°C to increase its temperature to
30°C ?
Calorimetry
1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes
This is also referred to as the Law of Conservation of Energy
Conservation of Energy
If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa
Calorimetry
Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system
Because of the Law of Conservation of Energy,the energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.
Calorimetry
Hsurroundings = - Hsystem
(m×C×T)surroundings = - (m×C×T)system
Calorimetry also uses the principle of thermal equilibrium.
Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature
Calorimetry
An example of CalorimetryExample:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water
at 24°C water. If the final temperature of the water is 32.5°,what is the specific capacity of
the metal?
Metal:m = 23.8 gTi = 100.0°CTf = 32.5°CCm = ? Water:m = 50.0 gTi = 24°CTf = 32.5°CCw = 4.184 J/g°C
An example of CalorimetryExample:
A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water
at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of
the metal?
Step 1: Find the Heat (ΔH) of the waterΔH = mCwΔT
ΔH = (50)(4.184)(32.5 – 24) ΔH = 1778.2 JStep 2: Remember that the water is the “surroundings”. So, the heat gained by the water is the same as the heat lost by the metal but opposite in sign!!
Step 3: Since you know ΔH of the metal system (-1778.2 J), use this the find Cm.
ΔH = mCmΔT
Now use the values for the metal!
-1778.2 = (23.8)Cm(32.5 – 100)-1778.2 = (23.8)(-67.5)Cm
(23.8)(-67.5)Cm = 1.11 J/goC
Let’s PracticeExample:
A 10.0 g of aluminum at 95.0°C is placed in a container of 100.0 g of water (C = 4.184
J/g°C) at 25.0°. The metal and water reach a final temperature of 26.5 oC. What is the
specific heat of the metal?