PES 2130 Fall 2014, Spendier
Lecture 2/Page 1
Lecture today: Chapter 18
1) Heat’s effect: thermal expansion and the absorption of heat by solids and liquids
2) Heat transfer
Announcements:
- HW 1 given out today
- No class or office hours next week Monday (Labor Day)
Last lecture:
Thermodynamics: The study of thermodynamics involves a collection of objects,
usually atomic in size, and how they behave as a system.
Phases (State of Matter) of Matter - Solid, liquid, or gas
Temperature: A measure of the average (translational) kinetic energy of the molecules
Temperature Scales and Conversions: Celsius, Fahrenheit, and Kelvin temperature
scales
Zeroth law of thermodynamics: If C is initially in thermal equilibrium with both A and
B, then A and B are in thermal equilibrium with each other.
Heat Q [J]: The transferred energy is called heat and it spontaneously flows from higher
to lower temperature objects.
Heat is positive when heat is absorbed by the system.
Heat is negative when heat is released or lost by the system.
Relevant concept to solve problems:
When heat flow occurs between two or more objects that are isolated from their
surroundings, the algebraic sum of the quantities of heat transferred to all bodies is zero:
Heat’s Effect
Because the internal energy includes more than just kinetic energy (also includes
potential energy), heat can have more than one effect on an object.
Heat can have three different effects on objects:
1) Thermal Expansion.
2) Change in temperature.
3) Change in phase.
PES 2130 Fall 2014, Spendier
Lecture 2/Page 2
1) Thermal Expansion: Change in length (volume) due to temperature changes.
a) We can model atoms in a solid as being held together by "springs" that are easier to
stretch than to compress.
b) As the energy increases the atoms oscillate with greater amplitude and the average
distance increases.
Therefore, increasing the temperature of a rod causes it to expand.
The change in length is given by
∆L = α L0 ∆T,
where α is the coefficient of linear expansion of the material. L0 is the length of the object
at reference temperature. ∆T=Tf -T0 is increase in temperature above reference
temperature T0.
If an object has a hole in it, the hole also expands with the object, as shown in the figure
at right. The hole does not shrink.
The change in volume due to thermal expansion is given by
∆V = ß V0 ∆T,
where ß = 3 α is the coefficient of volume expansion. V0 is the volume of the object at
reference temperature. ∆T is increase in temperature above reference.
Thermometers: We use the change in volume of mercury (Hg) in a thin container as a
thermometer.
PES 2130 Fall 2014, Spendier
Lecture 2/Page 3
Demo: Bi-metal Strips
Example 1:
By how much will a 4.0 m concrete slab expand going from 0ºC (freezing) to 38ºC
(100ºF)?
∆L = α L0 ∆T = (12*10-6/ ºC)(4.0 m)(38 ºC -0 ºC) = 1.8 mm.
This is enough to cause problems and hence you typically see spacers in between large
concrete slaps like bridges.
Exception to linear expansion (increasing volume with temperature)
- Thermal expansion of Water
- Between 0°C and 4°C, water decreases in
volume with increasing temperature.
- Water reaches is maximum density at 4°C.
- Because of this anomalous behavior, lakes
freeze from the top down instead of from the
bottom up. The entire lake must reach 4°C
before it will start to freeze.
2) Change in temperature.
A pot of water on a stove increases in temperature because of the transfer of heat Q from
the hot plate. By how much the temperature any substance increases due to heating
depends on
i) the mass m of the object and
ii) the material it is made out of.
The change in temperature is quantified by the specific heat, c of a material of mass m.
Q = mc ∆T = mc (Tf - Ti)
units for: c 
 J 
Q
 

mT
 kgK 
(1 cal = 4.1868 J)
PES 2130 Fall 2014, Spendier
Some examples:
Lecture 2/Page 4
copper: c = 386 J kg-1 K-1 (c = 0.0923 cal g-1 K-1)
glass: c = 840 J kg-1 K-1 (c = 0.20 cal g-1 K-1)
water: c = 4187 J kg-1 K-1 (c = 1.00 cal g-1 K-1)
It turns out that water has a higher specific heat than most substances. This explains why
most coastal locations have mild winters and summers.
Example 2:
Before going in for his annual physical, a 70.0 kg man whose body temperature is 37.0 ºC
consumes an entire 0.355 L can of a soft drink (mostly water) at 12.0 ºC
a) What will his body temperature be after equilibrium is attained? Ignore any heating by
the man’s metabolism. The specific heat of the man’s body is 3480 J kg-1 K-1 and of water
is 4190 J kg-1 K-1.
PES 2130 Fall 2014, Spendier
Lecture 2/Page 5
b) Is the change in his body temperature great enough to be measured by a medical
thermometer?
3) Change in phase. (hand boiler demo)
When you take a piece of ice out of the freezer and put it on the kitchen counter, the ice
will melt. This is a phase change, the transition from solid to liquid (or any phase to
another) and there is no change in temperature during a change in phase.
Plotting temperature as we add heat Q continuously to the system at constant pressure:
different "L"
for solid-liquid
as compared to
liquid-gas
different "c" for
steam as compared to
water
This is quantified by the heat of transformation (latent heat), L of a material.
Q=mL
units of L = [J/kg]
Note: The heat of fusion, Lf, is the heat per unit mass that is transferred in a solid-liquid
phase change.
The heat of vaporization, Lv, is the heat per unit mass transferred in a liquid-gas phase
change.
PES 2130 Fall 2014, Spendier
Lecture 2/Page 6
As we can see from the table, in general it takes a lot more energy to break a liquid into a
gas, than it does to break a solid into a liquid.
Example 3:
A glass contains 0.25 kg of a soft drink (mostly water) initially at 25 ºC. How much ice,
initially at -20 ºC, must be added to obtain a final temperature of 0.0 ºC with all the ice
melted? Neglect the heat capacity of the glass. The specific heat of water is
4190 J kg-1 K-1 and of ice is 2100 J kg-1 K-1. Heat of fusion to melt ice is 333 kJ/kg.