Matter and Energy
Drill
What mode of heat transfer best describes
the scenarios below:



Energy transfer through a brick wall
Air blowing over a hot surface
The sun warming your face on a cold day
Matter and Energy
Thermodynamics - “the study of systems and energy transfer”
Heat (Q) – energy transfer caused by a temperature
difference
Conduction
T

Q cond  - kA
x
Convection
Radiation

Q
conv  hA T
  AT 4
Q
rad
 t
QQ
Matter and Energy
Energy is transferred from a warm room at 20C inside a house to the outside
air at -10C through a single-pane window. The glass is 5mm thick with an
area of 0.5 m2 and a conductivity of 1.4 W/m K. What is the rate of heat
transfer through the glass?
Known: Heat flows through a window pane (A=0.5m2, ∆x = 5 mm, k = 1.4 W/m K)

Q
Find:
cond , [W]

Q
cond
Sketch:
Troom = 20C
= 293 K
Solution:
∆x
Assumptions: Closed System
Toutside = -10C
= 263 K
T

Q cond  - kA
x
2 293K  263K

Q

(1.4
W/m
K)(0.5
m
)
cond
1m
5mm( 1000
)
mm

Q
cond  - 4200 W
Matter and Energy
Cold air at -10C blows over a warm window-pane with a surface
temperature of 12C. The glass has a surface area of 0.5 m2 and the
convective heat transfer coefficient is h =100 W/m2 K. What is the rate of
heat transfer through the glass?
Known: Heat flows through a window pane (A=0.5m2, 100 W/m2 K)
 , [W]
Q
Find:
conv
Assumptions: Closed System
Toutside = -10C
Sketch:
= 263 K

Q
conv
Tglass = 12C
= 285 K
Solution:

Q
conv  hA T
2
2

Q

(100
W/m
K)(0.5
m
)( 285K  263K )
conv

Q
conv  1100 W
Matter and Energy
A small light bulb with a surface area of 0.025m2 and an emissivity of 0.6
fluoresces at a temperature of 100C. What is the rate of heat transfer from
the light bulb?
Known: Heat radiates from a bulb (A=0.025m2, Tbulb = 100 W/m2 K)
 , [W]
Q
Find:
rad
Assumptions: Closed System
Sketch:

Q
rad
Tbulb = 100 C
= 373 K
Solution:
  AT 4
Q
rad
  (0.6)(5.67x10 -8
Q
rad
  16.5 W
Q
rad
W
m 2 -K 4
)(0.025 m 2 )(373 K ) 4
Matter and Energy
Thermodynamics - “the study of systems and energy transfer”
A piston cylinder containing a gas was compressed over a period of 20 seconds.
The change in energy of the system was 300 kJ and the average rate of heat
transfer from the cylinder was 12 kW. Find the work done on the gas in kJ.
Known: A gas is compressed inside of a piston-cylinder
Find: W, [kJ]
Sketch:
Assumptions: Closed System

Q
Solution:
E1-2  Q1-2 - W1-2
W1-2  - E1-2  Q1-2
 t - E
W1-2  Q
W1-2  - 300 kJ  (-12 kJ/s)(20s)
W1-2  - 540 kJ
Matter and Energy
A vertical piston-cylinder device (D = 5 cm) contains a trapped mixture of gases.
The walls of the cylinder are insulated, but heat is allowed to enter through the
floor of the cylinder at a rate of 2.75 W over 20 seconds. The absolute gas
pressure is 101.3 kPa, initially. If the change in energy of the system is 41 J, how far
does the piston rise (in cm) during this process?
Known: A gas mixture is heated and expands inside of a piston-cylinder
Find:
rise of piston, h [cm]
Sketch:
Assumptions: Closed System ,
Vertical-piston cylinder -> Isobaric
Solution:
V1-2  r 2 h
h
πr 2V1-2
h
3
0.00013
πr 2 8 m
h
π(0.025 m) 2
h  0.07 m  7 cm
W1-2  P W
(V  V )
V1-2  12-2 1
P 1-2
W1-2  PV
14J
W

V

1
2
V11--22  101.3
P kPa
V1-2  0.000138 m 3

E
W11--22  Q11--22 --W
E1-22 Q  Qt 
Q(2.75
Q
)(t20 s)
Q

W
 55J 41J
WW
1- 21- 2 Q1- 2 - E1- 2
W1-2  14J
Q  (55 J)