What are common results of heat transfer?
1. Growth of temperature.
2. A phase transition (melting ice).
3. Mechanical work.
In cases #2 and #3 there may be NO temperature variation.
Case #1, no phase transition or work done.
How much does the temperature vary?
Heat capacity of the object C, measured in J/K;
tells you how much Joules of heat you need to transfer to
increase the temperature of the object by 1 K (or 1 ºC).
!Q
C=
!T
!Q = C!T
Heat is energy in transit! Positive, when obtained.
Heat capacity of the object C, measured in J/K;
tells you how much Joules of heat you need to transfer to
increase the temperature of the object by 1 K (or 1 ºC).
!Q
C=
!T
The term “heat capacity” is used for historic reasons and is confusing!
It sounds as if the object “contains heat”, whereas by definition heat is
energy in transit. An object contains some internal energy (not heat!) and
its temperature is a measure of this internal energy.
An analogy from mechanics: you do
some work on an object and the
object gains the same amount of
energy (kinetic, potential) as a result.
Mechanical work also relates to a
process, not state.
!K + mg!h = W
Heat capacity is an extensive (integral) parameter:
When you bring two objects together, heat capacity of the system of the
two objects becomes the sum of the two individual heat capacities.
It is convenient to introduce specific heat, c, which
is heat capacity of a material per unit mass.
Specific heat is measured in J/(K⋅kg).
C = cm
C
!Q
c= =
m m!T
!Q = cm!T
Heat capacity of a water balloon is large due to both high
specific heat and large mass of the water inside the balloon.
Specific heat, c, is heat capacity of a material per unit mass.
Specific heat is measured in J/(K⋅kg).
!Q = cm!T
C
!Q
c= =
m m!T
The equilibrium temperature.
Situation: two objects with different temperatures, T1 and T2, are
brought in a thermal contact and reach thermal equilibrium at a
temperature T after a while.
Heat ΔQ1 is transferred to Object 1; heat ΔQ2 is transferred to Object 2.
By energy conservation:
!Q1 + !Q2 = 0
!Q1 = "!Q2
By definition of heat capacity and specific heat:
"Q1 = C1"T1 = m1c1"T1 = m1c1 (T ! T1 )
"Q2 = C2 "T2 = m2 c2 "T2 = m2 c2 (T ! T2 )
m1c1 (T ! T1 ) + m2 c2 (T ! T2 ) = 0
m1c1T1 + m2 c2T2 C1T1 + C2T2
T=
=
m1c1 + m2 c2
C1 + C2
Heat transfer
1.Conduction.
2. Convection
3. Radiation
Setting:
A rectangular slab of thickness Δx
and with an area A.
The front side of the slab is at a
temperature T;
the back side has a somewhat
different temperature, T+ΔT.
We are trying to calculate the heatflow rate, the amount of heat flowing
through the slab per unit time,
H = ΔQ/Δt.
We expect H to be proportional to the area of the slab, the
temperature difference, ΔT, between the back and the front and
inversely proportional to the thickness of the slab, Δx.
H should also somehow depend of properties of the material the slab is
made of…
Bringing all the parts together:
!T
H = "kA
!x
The coefficient k reflects specific
properties of the material of the
slab and is called thermal
conductivity
H = ΔQ/Δt – heat-flow rate is measured in Joules/second, J/s, or Watts,
W.
Thermal conductivity, k, is measured in W/(m⋅K).
Thermal conductivities of different materials.
Best heat conductor – Copper; use it when you build heat sink, as a
material for pipes in your cooling system, a radiator.
Worst heat conductors are the best insulating materials – air,
fiberglass (layers in the walls of houses in cold regions), styrofoam
(cups for your hot coffee).
Heat-flow rate equation
!T
!T
H = " kA
="
!x
!x /(kA)
Is similar to the Ohm’s law:
V
I=
R
The current, I = Δq/Δt,
amount of charge per unit time,
is analogous to the heat-flow rate, H = ΔQ/Δt.
The voltage, V, the factor driving the electric
current, is analogous to temperature difference, ΔT.
Continuing the analogy: electric resistance, R, is analogous to
Thermal resistance is introduced as
Thermal resistance is introduced as
Heat-flow rate equation
!T
H ="
R
A composite slab is analogous to two
resistors connected in a series.
Heat-flow rate
"Ttotal T3 ! T1
H =!
=
Rtotal
R1 + R2
Reducing heat-flow rate for better thermal insulation.
Better thermal insulation… red or blue?
Licking an ice cream, which is
frozen, seems to be OK…
What about the door handles, when it is freezing outside?
Which one would
you prefer?
Cold metals are especially bad because of
their high… thermal conductivity
In order to keep your tongue above 0 °C you basically
have to heat the whole piece of metal…
Otherwise…
Thermal conductivities of different materials.
Sample problem
A home heating system supplies heat at the maximum rate of 40 kW.
If the house loses 1.1 kW for each °C between inside and outside,
what is the minimal outdoor temperature for which the heating system
can maintain 20 °C inside?
Solution: the lowest temperature outside for which the heating
system can maintain 20 °C inside corresponds to the case, when the
maximal power of the heater, Hmax=40 kW all flows outside because of
temperature difference.
H out
1.1kW
=
!T = H max = 40 kW
1°C
40 kW
!T =
°C = 36 °C
1.1 kW
Tout = Tin ! "T = !16 °C
Convection - Heat transfer in a gas or liquid by the
circulation of currents from one region to another.
Can be forced or spontaneous (natural).
Hot and cold liquid is brought in a thermal contact;
it reduces the distance across which the
conduction occurs and increases the contact area.
!T
H = " kA
!x