Physics 7B
Fall, 2006
Homework 3 Solutions
19.74
The rate of thermal energy flow is the same for the brick and
the insulation:
Δ
Q
/
Δ
t
=
A
(
T
1
–
T
2
)/
R
eff
=
A
(
T
1
–
T
int
)/
R
2
=
A
(
T
int
–
T
2
)/
R
1
,
where
T
int
is the temperature at the brickinsulation interface.
By equating the first term to each of the others, we have
R
eff
(
T
1
–
T
int
) =
R
2
(
T
1
–
T
2
), and
R
eff
(
T
int
–
T
2
) =
R
1
(
T
1
–
T
2
).
If we add these two equations, we get
R
eff
(
T
1
–
T
2
) =
R
1
(
T
1
–
T
2
) +
R
2
(
T
1
–
T
2
), which gives
R
eff
=
R
1
+
R
2
.
This shows the usefulness of the
R
value.
For the
R
value of the brick we have
R
1
=
L
1
/
k
1
= [(4.0 in)/(12 in/ft)](3.28 ft/m)/(0.84 J/s
·
m
·
C°)(1 Btu/1055 J)(
3600 s /1 h)(5 C°/9 F°)
= 0.69 ft
2
·
h
·
F°/Btu.
Thus the total
R
value of the wall is
R
eff
=
R
1
+
R
2
= 0.69 ft
2
·
h
·
F°/Btu + 19 ft
2
·
h
·
F°/Btu = 19.7 ft
2
·
h
·
F°/Btu.
The rate of heat loss is
Δ
Q
/
Δ
t
=
A
(
T
1
–
T
2
)/
R
eff
= [(240 ft
2
)(10 F°)/(19.7 ft
2
·
h
·
F°/Btu)](1055 J/Btu)/(3600 s/h) =
36 W
.
19.90
(
a
) We find the rate of heat flow through the clothing from
Δ
Q
/
Δ
t
=
kA
(
Δ
T
/
L
)
= (0.025 J/s
·
m
·
C°)(1.9 m
2
)[34°C – (– 20°C)]/(0.035 m) =
73 W
.
(
b
) When the clothing is wet, we have
Δ
Q
/
Δ
t
=
kA
(
Δ
T
/
L
)
= (0.56 J/s
·
m
·
C°)(1.9 m
2
)[34°C – (– 20°C)]/(0.0050 m) =
1.1
×
10
4
W
.
19.101
.
The heat of fusion from the ice forming at the bottom of the ice must be conducted through the ice to
the air. When the ice has a thickness
x
, we have
d
Q
/d
t
=
kA
Δ
T
/
x
=
ρ
(
A
d
x
/d
t
)
L
f
.
We separate the variables and integrate:
k
!
T
"
L
f
d
t
0
t
=
x
d
x
0
D
; whichgives
k
!
T
t
"
L
f
=
D
2
2
.
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 Fall '08
 Packard
 Energy, Work, Heat, Thermal Energy, Ideal gas equation, Adiabatic process, Isothermal process, Reff

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