Chapter 14: Heat
Chapter Outline
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Heat as energy transfer
Temperature vs. heat vs. internal energy
Internal energy of an Ideal Gas
Specific Heat
Calorimetry
Latent Heat
Heat Transfer: conduction, convection, and
radiation
Heat is a Transfer of energy
• Heat always flows from hot to cold.
• In the 18th century scientists believed heat
was an actual, physical thing that flowed from
one object to another.
– They pictured it as a fluid and named that fluid
caloric
• The scientists were never able to detect this
fluid.
The New Model
• Scientists never discovered caloric because
heat is not a physical object.
• In the 1800s several scientists worked on a
new model of heat, one such scientist was an
English brewer named James Prescott.
• Prescott filled a vessel with water that had
paddles in it that would move when a mass
was dropped.
The Result
• What Prescott discovered was that the
temperature of the water increased.
• Prescott showed that mechanical energy is
transferred from one object to another as
heat.
• What we have been calling the moss-pit is the
transfer of heat.
Units of Heat
• calorie: the energy needed to heat 1g of water
1C
• Calorie (aka Kilocalorie aka food calorie): the
energy needed to heat 1kg of water 1C
• 1 calorie = 4.186J
• 1 kilocalorie(kcal) = 4186J
• These bottom 2 conversions show how heat IS
energy
Please Consider the Following
• If heat is energy and work is the exchange of
energy, then can heat do work?
Example 1
• How tall a flight of stairs would a person have
to climb to burn off 500 Calories? Assume the
person is 60kg.
Solution
• Step 1 – convert Cal to J
• 500kcal (4186J/kcal) = 2.1E6J
• So 500 Calories gives us 2.1E6J of work, how
high will that work take us?
• Step 2 W = mgh
• h = W/mg = 2.1E6J / (60kg x 9.8m/s2) =
3600m or over 11,000 ft!
Conversation of Energy
• Heat factors into the conversation of energy
• If any kinetic energy is lost, it is lost as heat.
• KEi = KEf + Q, where Q is heat, (why Q? I don’t
know)
Example
• A 3.0g bullet travelling with a speed of 400m/s
passes through a tree and slows down to
200m/s. How much heat, Q, is produced and
shared by the bullet to the tree?
Solution
• Step 1: convert to kg
• 3.0g = 3.0E-3kg
• Step 2: KEi = KEf + Q
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Q = KEi - KEf
Q = 1/2mvi2 – 1/2mvf2
Q = 1/2m(vi2 – vf2)
Q = 180J = 43cal
Temperature, Heat, and Internal
Energy
• The sum total of all the energy of all the
molecules in an object is called the internal
energy.
• Remember, temperature is the average kinetic
energy of all molecules in an object.
Consider This
• If you touch a glass of water that is the same
temperature as your hand is there heat
transfer?
• Does the water have internal energy?
Internal Energy of an Ideal Gas
• The internal energy of an ideal gas, U,
depends only on the temperature of the gas
and how many moles of gas there are.
• U = 3/2nRT
• Again this is for an ideal, monatomic gas.
• For a real gas, rotational and vibrational
energies would come into play.
Heat and Temperature
• As heat is put into an object the temperature
goes up.
• But by how much?
• Well that depends…
Some questions to ask
• Is there a difference to how long it takes to
boil a pot of water if it is a little pot or a big
pot?
• Does it take more or less energy to get heavier
molecules moving?
• Will it take more heat to get to a higher
temperature?
The math
• Q = mcΔT, where m is mass, T is temperature,
and c is called specific heat and is different for
every element.
• c = Q/mΔT and its units are J/kgCo
Heat and conservation of energy
• Imagine a completely isolated system where
no energy can flow into or out of the system.
• In such a system the energy must be
conserved.
• So, if heat is lost by one part of the system it
must be gained by another part.
Calorimetry
• Heat lost = heat gained
• Remember from yesterday, Q = mcΔT
• So mc(Tf – Ti) = mc(Tf – Ti)
Example
• If 200cm3 of tea at 950C is poured into a 150g
glass cup initially at 250C, what will be the
final temperature T of the mixture when
equilibrium is reached, assuming no heat
flows to the surroundings?
Solution part 1
• We need to find m, c, and ΔT
• m: mass is density times volume so
m = 200E-6m3 x 1E3kg/m3 = 0.20kg
• c = 4186J/kgC (because tea is basically water)
• Conservation of energy gives us
mteactea(950C – T) = mcupccup(T – 250C)
Solution part 2
• mteactea(950C – T) = mcupccup(T – 250C)
• (0.20kg)(4186J/kgC)(95 – T) =
(0.15kg)(840J/kgC)(T – 25)
• Solving for T gives us T = 890C
Finding Specific Heats
• What could you do to find the specific heat of
an unknown substance?
What scientists do
• They perform what is called calorimetry
– They heat the object to a certain temperature.
– They quickly place the hot object into an amount
of water whose mass and temperature are known.
– They record the final temperature of the water to
see how much energy was transferred.
• Important, when doing this, scientists try to
keep the system well insulated from the
outside environment.
The details
• Heat lost by sample = heat gained by water +
heat gained by the container
• msamplecsampleΔTsample = mwatercwaterΔTwater +
mcupccupΔTcup
Example
• We want to know the specific heat of a new
metal alloy that we created. A 0.150kg sample
of the new alloy is heated to 5400C. It is then
placed in 400g of water at 100C, which is
contained in a 200g aluminum calorimeter
cup. If the final temperature of the water is
30.50C, what is the specific heat of our alloy?
Solution
• msamplecsampleΔTsample = mwatercwaterΔTwater +
mcupccupΔTcup
• (0.150kg)(csample)(540 – 30.5) =
(0.400kg)(4186J/kgC)(30.5 – 10) +
(0.200kg)(900J/kgC)(30.5 – 10.0)
• 76.4csample = (34,300 + 3700)J/kgC
• csample = 500J/kgC
Bomb calorimeter
• A bomb calorimeter is used to measure the heat
released when a substance burns or explodes.
• It is used to find the calorie content of foods or
the energy released by a type of explosive.
• A carefully weighted sample of the substance,
together with an excess of oxygen at high
pressure, is placed in a sealed container (the
bomb) which is placed in the water and ignited.
Conduction
• Heat transfer by molecules colliding
• The flow of heat is related to the difference in
temperature
Q
t
 kA
T1  T 2
l
• Where l is the distance between the ends of
the two objects, A is the area, and k is called
the thermal conductivity, which is different for
different materials
Conductors vs. Insulators
• Materials that have a high k transfer heat
quickly and are called conductors.
• Materials that have a low k transfer heat
slowly and are called insulators.
Vive la Resistance
• Insulators commonly have an R value assigned
to them to illustrate how good an insulator
they are.
• The higher the R, the better the insulator
• R = l/k where l is the thickness of the material
and k is its thermal conductivity
Convection
• Heat transfer by the mass movement of
molecules from one place to another.
• 2 types
– Forced: like a furnace blowing hot air into a room
– Natural: warm air rises
Radiation
• Heat transfer that requires no medium at all.
• This is how the sun transfers its heat to earth
or how IR lamps keep food warm
Stefan-Boltzmann equation
Q
t
 e  AT
4
Where σ is called the Stefan-Boltzmann
constant, σ = 5.67E-8W/m2K4 and e is
called the emissivity and is between 0
and 1
More on emissivity
• Dark objects like black clothing or dark roofs
have an emissivity close to 1 and absorb
radiation
• Light objects like white roofs reflect radiation
Going Tanning
• The sun sends 1350J of energy per second per
square meter at a right angle to the earth.
• About 1000W/m2 reaches the surface.
• The following equation can be used to find
how much radiation an object absorbs from
the sun.
Q
t
 (1000 W / m ) eA cos 
2