This document discusses heat transfer through conduction. It defines key terms like heat capacity, specific heat, and latent heat. It explains that heat transfer occurs as higher kinetic energy molecules collide with and transfer energy to lower kinetic energy molecules. The rate of conductive heat transfer depends on the temperature difference, contact area, and thermal conductivity of the materials. An example calculation shows how to determine the temperature at the junction and heat transfer rate when heat is conducted through two layers with different thermal conductivities.

1. Lecture 24 Purdue University, Physics 220 1
Lecture 24
Heat
PHYSICS 220
Lecture 24 Purdue University, Physics 220 2
Heat Capacity and Specific Heat
Heat capacity =Q/!T
• shows how much heat is required to change the
T of object (system)
• Specific heat c = Q/m!T
• Q = c m !T
Heat required to increase temperature depends on
amount of material (m) and type of material
• Heat adds energy to object/system
• IF there is no dissipation then:
Heat increases internal energy: Q = !U
Heat increases temperature: Q = C !T
Lecture 24 Purdue University, Physics 220 3
Exercise
After a grueling work out, you drink a liter of cold
water (0 C). How many Calories does it take for
your body to raise the water up to body
temperature of 36 C?
A) 36 B) 360 C) 3,600 D) 36,000
1 liter = 1,000 grams of H20
1000 g x 1 calorie/(gram degree) x (36 degree) = 36,000 calories
36,000 calories = 36 Calories!
Lecture 24 Purdue University, Physics 220 4
Question
Suppose you have equal masses of aluminum and
copper at the same initial temperature. You add 1000 J
of heat to each of them. Which one ends up at the
higher final temperature
A) aluminum
B) copper
C) the same
!T = Q/cm
Substance c in J/(kg-C)
aluminum 900
copper 387
iron 452
lead 128
human body 3500
water 4186
ice 2000

2. Lecture 24 Purdue University, Physics 220 5
Specific Heat for Ideal Gas
• Monatomic Gas (single atom)
Translational kinetic energy only
At constant Volume work = 0
Q = !U = 3/2 nR!T
CV = 3/2 R = 12.5 J/(K mole)
Cv – specific heat at constant volume.
• Diatomic Gas (two atoms)
Can also rotate
CV = 5/2 R = 20.8 J/(K mole)
Lecture 24 Purdue University, Physics 220 6
Phase Transitions
• A phase transition occurs whenever a material is changed
from one phase, such as the solid phase, to another
phase, such as the liquid phase.
– Phase transitions occur at constant temperature.
– The latent heat of vaporization LV is the heat per unit mass that
must flow to change the phase from liquid to gas or from gas to
liquid.
• Fusion occurs when a liquid turns into a solid.
• Evaporation occurs when a liquid turns into a gas.
• Sublimation occurs when a solid changes directly to a gas
without going into a liquid form.
Demo 3B - 04
T pinned at boiling
point of water
which is below
ignition point
for cup
Lecture 24 Purdue University, Physics 220 8
• As you add heat to water, the temperature increases
for a while, then it remains constant, despite the
additional heat!
• Latent Heat L [J/kg] is heat which must be added (or
removed) for material to change phase (liquid-gas).
• |Q| = m L
Latent Heat
T
Q added to water
water
temp
rises
water
changes
to steam
(boils)
steam
temp
rises
100oC
Latent Heat
Substance Lf (J/kg) Lv (J/kg)
water 33.5 x 104 22.6 x 105
f=fusion v=vaporization

3. Why
When boiling the most energetic molecules escape
Reducing the average T
But heat added to continue boiling so
process goes to completion
Lecture 24 Purdue University, Physics 220 10
Phase Diagram
H2O
Lecture 24 Purdue University, Physics 220 11
Phase Diagram
CO2
Lecture 24 Purdue University, Physics 220 12
Exercise
During a tough work out, your body sweats (and evaporates)
1 liter of water to keep cool (37 C). How much cold water
would you need to drink (at 2 C) to achieve the same thermal
cooling? (recall CV = 4.2 J/g for water, Lv=2.2x103 J/g)
A) 0.15 liters B) 1.0 liters C) 15 liters D) 150 liters
Qevaporative = L m = 2.2x103 kJ/kg x 1kg
Qc = c m !t = 4.2kJ/kgK x 35K x m
m = 2.2x103 / 147 = 15kg or 15 liters!

4. Lecture 24 Purdue University, Physics 220 13
Boiling Point
Going from Lafayette to Denver the
temperature at which water boils:
A) Increases B) Decreases C) Same
Lecture 24 Purdue University, Physics 220 14
Exercise
How much ice (at 0 C) do you need to add to 0.5 liters of a
water at 25 C, to cool it down to 10 C?
(L = 80 cal/g, c = 1 cal/g C)
Qwater
= mc!T
= (0.5kg)(1cal / gC)(15C)
= (7,500 calories)
Qice
= mL + mc!T
" m =
Qice
L + c!T
=
7,500cal
80cal / g + (1cal / gC)(10)
= 83.3 grams
Not same m
Lecture 24 Purdue University, Physics 220 15
Exercise
Ice cube trays are filled with 0.5 kg of water at 20 C and
placed into the freezer. How much energy must be
removed from the water to turn it into ice cubes at -5 C?
(L = 80 cal/g, cwater = 1 cal/g C, cice = 0.5 cal/g C)
Q1
= mcwater
!T1
= 500 "1" (#20) = #10000(cal)
Water going from 20 C to 0 C:
Water turning into ice at 0 C: Q2
= !mL
= !500 " 80 = !40000(cal)
Ice going from 0 C to -5 C: Q3
= mcice
!T2
= 500 " 0.5" (#5) = #1250(cal)
! Q = Q1
+ Q2
+ Q3
= "51250(cal)
Heat transfer
• Conduction
• Convection
• Electro-magnetic radiation
Lecture 25 Purdue University, Physics 220 16

5. Demo 3B - 03
Char were little conduction
Lecture 25 Purdue University, Physics 220 18
Heat Transfer: Conduction
• Hot molecules have more KE than
cold molecules
• High-speed molecules on left
collide with low-speed molecules on
right
– energy transferred to lower-speed
molecules
– heat transfers from hot to cold
– vibrations
Lecture 25 Purdue University, Physics 220 19
Heat Transfer: Conduction
• I = rate of heat transfer = Q/t [J/s]
I = " A (TH-TC)/d
• Q/t = " A !T/!x
" = thermal conductivity
• Units: J/s*m*C
• good conductors…high ", e.g., metal
• good insulators … low ", e.g., plastic
R = d/(A") = thermal resistance
TH
Hot
TC
Cold
d = !x
Area A
Lecture 25 Purdue University, Physics 220 20
Conduction
Which of the following is an example of conductive heat
transfer?
A) You stir some hot soup with a silver spoon and notice that
the spoon warms up.
B) You stand watching a bonfire, but can’t get too close
because of the heat.
C) Its hard for central air-conditioning in an old house to cool
the attic.

6. Lecture 25 Purdue University, Physics 220 21
• Find I=Q/t in J/s
Key Point: Continuity (just like fluid flow)
• I1 = I2
• "1A(T0-TC)/!x1 = "2A(TH-T0)/!x2
• solve for T0 = temp. at junction
• then solve for I1 or I2
• TH-T0 = I R1 and T0-TC = I R2
!T = (TH-T0) + (T0-TC) = I (R1 + R2)
!x1 = 0.02 m A1 = 35 m2 k1 = 0.080 J/s*m*C
!x2 = 0.075 m A2 = 35 m2 k2 = 0.030 J/s*m*C
answer: T0=2.27 C I=318 Watts
Inside: TH = 25C
Outside: TC = 0C
I1 I2
T0
Conduction with 2 Layers