Temperature, Heat,
& Combustion
EGR 1301: Introduction to
Engineering
EGR 1301
Models
• “A system of postulates, data, and inferences
presented as a mathematical description of an
entity or state of affairs”
• Quantitative approximation of reality
Source: Merriam-Webster .com, 2010
 Mathematical equations
 Computer simulations
 Physical scale models
• Why do we use them?
 Reality is too complex!!!
EGR 1301
Energy Conversion Tables
“For those who want some proof that physicists are human, the proof is in
the idiocy of all the different units which they use for measuring energy.”
Richard Feynman
Source: Foundations of Engineering, Holtzapple & Reece, 2003
EGR 1301
Utility of Energy for Analysis
Source: http://www.fullspectrumsolutions.com/26w_powercompact_65_prd1.htm
• Incandescent bulb
 Resistance heating in
filament  Light
• When filament reaches
sufficiently high
temperature  Light is
radiated
• Fluorescent bulb
 Stream of electrons
collide with Hg
electronsLight
• Generates very little
heat
 60 Watts of electricity
 23 Watts of electricity
• 800 lumens of light
• ~10 cals of heat
• 800 lumens of light
• ~1 cal of heat
EGR 1301
Temperature – What Is It?
• NOT the same as HEAT
• A quantitative measure of “hotness”
• More accurately described on an atomic
scale
 Measures vibrational kinetic energy
EGR 1301
Temperature vs. Heat
• Temperature
 A measure of the intensity of internal energy
in a system (gas, liquid, or solid)
• Heat
 A measure of the total quantity of thermal
energy flow into or out of a system
EGR 1301
Temperature vs. Heat
• Example:
 A cup of water at 60°C has much less energy
than a hot water heater full of water at 60°C.
 BUT, the intensity of heat is the same.
EGR 1301
Heat Capacity
• Energy required to raise temperature of
matter by one degree (at constant
pressure or constant volume)
Q
C
mT
 Q = energy in calories
 m = mass in grams
 ΔT = temperature change in degrees (C or K)
EGR 1301
Constant Pressure
Heat Capacities
Source: Foundations of Engineering, Holtzapple & Reece, 2003
EGR 1301
Converting Work into Heat:
Joule’s Experiment
W  Fx
F  ma
Q  CmT
U  Q
Source: Foundations of Engineering, Holtzapple & Reece, 2003
W Q
Ek  E p  U  Win  Wout  Qin  Qout  M in  M out
EGR 1301
Heat Capacity
Example Problem
• In Joule’s experiment,
 Beaker contains 5 kg of water
 Mass spinning the stirrer is 90 kg
(g=9.81m/s2)
 The water increases in temperature by 0.1°C
 How far did the mass travel?
Q  CmT
F  ma
W  Fx
W Q
EGR 1301
Heat Capacity
Q  CmT
W  Q F  ma
W  Fx
 1000 g 
1.00cal
5kg 
Q
0.1C   500cal
g  C
 1kg 


 kg  m
1



W Q
500cal
4.184 J 

s


x  


F ma  90kg  9.81m  1cal  1J





s 
2
2
x  2.4m
EGR 1301
2





States of Matter
Source: Foundations of Engineering, Holtzapple & Reece, 2003
EGR 1301
Phase Diagram
Source: Foundations of Engineering, Holtzapple & Reece, 2003
EGR 1301
Phase Change
• Constant temperature process of transition
between phases
 Melting / Solidification
 Boiling (vaporization) / Condensation
EGR 1301
Phase (or State) Change Energy
Qvap  mH vap
Q fus  mH fus
• Where
 m = mass (kg)
 ΔHvap = latent heat of vaporization (kJ/kg)
 ΔHfus = latent heat of fusion (kJ/kg)
EGR 1301
Phase-Change Energy
Source: Foundations of Engineering, Holtzapple & Reece, 2003
EGR 1301
Combustion
• Similar to phase change
Qcomb  mH comb
• Where
 Qcomb = energy released (MJ)
 m = mass (kg)
 ΔHcomb = specific heat of combustion (MJ/kg)
• Table 22.4
EGR 1301
Example 1:
Phase-Change Energy
• When water changes from solid to liquid, it
must absorb 333.56 kJ/kg from the
surroundings
 What is the energy absorbed to melt ice in
units of cal/g?
kJ  1kg  1000 J  0.2390cal 
333.56 



kg  1000 g  kJ 
1J

cal
 79.7
g
EGR 1301
Example 2a:
1st Law of Thermodynamics
• If you have 100 g of water at 22°C and add 20 g
of ice at 0°C, what will be the temperature of the
100 g of water once all the ice has melted to
form 20 g of water at 0°C?
Given :
m  100 g
m  20 g
T  22C
T  0C
1
1
2
2
EGR 1301
Example 2a:
1st Law of Thermodynamics
Given :
m  100 g m2  20 g
T  22C T2  0C
1
1
Eqns :
Q  CmT
H
fusion
 mH 
 1.00cal 
 79.7cal 

100 g T   20 g 

 g  C 
 g 
T  15.94C
T  22  15.94C  6.06C
1b
EGR 1301
Example 2b:
1st Law of Thermodynamics
• What will be the final temperature when the
system temperature is uniform (i.e., water from
melted ice has warmed and surrounding water
has further cooled so that all water is at one
temperature)?
Given :
m1  100 g
T1b  6.06C
m2  20 g
T2  0C
 1.00cal 

100 g (T  6.06C )
 g  C 
 1.00cal 

20 g (T  0C )
 g  C 
f
f
T  5.05C
f
EGR 1301
Example 3:
Latent Heat
• If the latent heat of vaporization for water is
2,256.7 kJ/kg, what is the latent heat for water in
cal/g?
cal
kJ  0.2390cal 
2256.7 
  539.4
kg 
1J
g

EGR 1301
Example 4a:
1st Law of Thermodynamics
• If by sweat and evaporation, you lose 0.031
slugs of water during exercising, how many
calories of energy in the form of heat is removed
from your body?
 Weight (mg) of 1 lb is associated with mass of 0.031
slugs
 Note on p.688 the conversion from slugs to grams
 14594 g  539.4cal 
0.031slug 


g
 slug 

 244032cal
EGR 1301
Example 4b:
1st Law of Thermodynamics
• If your body mass is 68,100 g (i.e., 150 lbs), how
much would your body temperature rise if you
did not sweat and evaporate the sweat in order
to cool yourself?
 Assume the heat capacity for your body is that for
water, because your body is ~ 75% water.
Q  CmT
 1cal 
244032cal  
(68100 g )T
 g  C 
 9 F 
T  3.58C 
  6.45 F
 5C 
EGR 1301