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Problem 1 ( 15+15 points)
We assume that the total heat flow is approximately the same as it would
be through a flat slab of area 0.80 m
2
and thickness 2.0(1.5) cm =
0.020 (0.015) m.
(a) The heat current H is then expressed by:
J/s.
12
W
12
m
0.020
C
0
C
30
)
m
K)](0.80
W/(m
010
.
0
[
2
−
=
−
=
°
−
°
−
=
−
−
=
L
T
T
kA
H
L
H
(15 points, the sign is not important so accept the answer with H>0)
(b) The total heat flow Q in one day (=86,400 s) is:
J.
10
1.0
s)
0
J/s)(86,40
12
(
6
×
=
=
=
Ht
Q
The quantity of ice melted by this quantity of heat is:
kg.
3.0
J/kg)
10
J)/(3.34
10
0
.
1
(
/
5
6
=
×
×
=
=
f
L
Q
m
(15 points)
PHY 126 Midterm Exam II
Chapters
15,16,17
J/s.
19
W
19
m
0.015
C
0
C
35
)
m
K)](0.80
W/(m
010
.
0
[
2
−
=
−
=
°
−
°
−
=
−
−
=
L
T
T
kA
H
L
H
(A)
(B)
J.
10
1.6
s)
0
J/s)(86,40
19
(
6
×
=
=
=
Ht
Q
kg.
4.8
J/kg)
10
J)/(3.34
10
6
.
1
(
/
5
6
=
×
×
=
=
f
L
Q
m
(A)
(B)
(A)
(B)
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View Full Document Problem 2 (30 points)
(a)From Bernoulli’s Eq. and use the facts that the velocity of the
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This note was uploaded on 05/28/2011 for the course PHY 126 taught by Professor Staff during the Summer '08 term at SUNY Stony Brook.
 Summer '08
 Staff
 Physics, Current, Heat

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