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Unformatted text preview: erature at a distance of 0.15 m from the hot end of the bar is T = 306 °C − 19 C° = 287 °C Chapter 13 Problems 693 10. REASONING The heat lost by conduction through the wall is Qwall and that lost through
the window is Qwindow. The total heat lost through the wall and window is Qwall + Qwindow.
The percentage of the total heat lost by the window is Qwindow
Percentage = × 100% Qwall + Qwindow (1) The amount of heat Q conducted in a time t is given by Q= ( k A∆T ) t (13.1) L where k is the thermal conductivity, A is the area, ∆T is the temperature difference, and L is
the thickness.
SOLUTION Substituting Equation (13.1) into Equation (1), and letting the symbols “S”
denote the Styrofoam wall and “G” the glass window, we have that Qwindow
Percentage = × 100% Qwall + Qwindow kG AG ( ∆T ) t
kG AG LG
LG × 100% = = kS AS ( ∆T ) t kG AG ( ∆T ) t kS AS + kG AG
+ L
LG
LS
LG
S × 100% Here we algebraically eliminated the temperature difference ∆T and the time t, since they
are the same in each term. According to Table 13.1 the thermal conductivity of glass is
kG = 0.80 J/ ( s ⋅ m ⋅ C° ) , while the value for Styrofoam is kS = 0.010 J/ ( s ⋅ m ⋅ C° ) . The
percentage of the total heat lost by the window is 694 THE TRANSFER OF HEAT kG AG LG
Percentage = kS AS + kG AG
L
LG
S × 100% ( ) 0.80 J/ ( s ⋅ m ⋅ C° ) 0.16 m2 −3 2.0 × 10 m
=
2
0.80 J/ ( s ⋅ m ⋅ C° ) 0.16 m2 0.010 J/ ( s ⋅ m ⋅ C° ) 18 m + −3
0.10 m 2.0 × 10 m ( ) ( ) × 100% = 97 % 11. REASONING To find the total heat conducted, we will apply Equation 13.1 to the steel
portion and the iron portion of the rod. In so doing, we use the area of a square for the cross
section of the steel. The area of the iron is the area of the circle minus the area of the square.
The radius of the circle is one half the length of the diagonal of the square.
SOLUTION In preparation for applying Equation 13.1, we need the area of the steel and
2
the area of the iron. For the steel, the area is simply ASteel = L , where L is the length of a side of the square. For the iron, the area is AIron = π R2 – L2. To find the radius R, we use the
Pythagorean theorem, which indicates that the length D of the diagonal is related to the
2
2
2
length of the sides according to D = L + L . Therefore, the radius of the circle is
R = D / 2 = 2 L / 2 . For the iron, then, the area is
2 AIron 2L 2
2 π
=πR −L =π − L = − 1 L
2
2 2 2 Taking values for the thermal conductivities of steel and iron from Table 13.1 and applying
Equation 13.1, we find Chapter 13 Problems 695 QTotal = QSteel + QIron ( kA∆T ) t ( kA∆T ) t π ( ∆T ) t
=
+
= kSteel L2 + kIron − 1 L2 L
L
2 L Steel Iron J
J 2
2 π = 14 ( 0.010 m ) + 79 − 1 ( 0.010 m ) s ⋅ m ⋅ C° s ⋅ m ⋅ C° 2 × ( 78 °C − 18 °C )(120 s ) =
0.50 m 85 J 12. REASONING
a. The heat Q conducted through the tile in a time t is given by Q = ( kA ∆T ) t L
(Equation 13.1), where k is the thermal conductivity of the tile, A is its crosssectional area,
L is the distance between the outer and inner surfaces, and ∆T is the temperature difference
between the outer and inner surfaces. b. We will use Q = cm∆T (Equation 12.4) to find the increase ∆T in the temperature of a
mass m = 2.0 kg of water when an amount of heat Q is transferred to it. SOLUTION
a. The time t = 5.0 min must be converted to SI units (seconds): ( ) 60 s
t = 5.0 min 1 min 2 = 3.0 × 10 s Because the tile is cubical, its thickness is equal to the length L of one of its sides, and its
crosssectional area A is the product of two of its side lengths A = (L)(L) = L2. Applying
( kA ∆T ) t (Equation 13.1), we obtain the amount of heat conducted by the tile in five
Q=
L
minutes: ( kA ∆T ) t = ( kL
Q=
L ( 2 ) ∆T t
L ) = ( kL ∆T ) t ( )( ) = 0.065 J s ⋅ m ⋅ Co ( 0.10 m ) 1150 o C − 20.0 o C 3.0 × 10 2 s = 2200 J 696 THE TRANSFER OF HEAT b. Solving Q = cm∆T (Equation 12.4) for the increase ∆T in temperature, we obtain
∆T = Q
cm (1) In Equation (1), we will use the value of Q found in part (a), and the specific heat capacity c
of water given in Table 12.2 in the text. With these values, the increase in temperature of
two liters of water is
∆T = ( 2200 J ) 4186 J/ kg ⋅ Co ( 2.0 kg ) = 0.26 Co 13. SSM REASONING AND SOLUTION The rate of heat transfer is the same for all three
materials so
Q/t = kpA∆Tp/L = kbA∆Tb/L = kwA∆Tw/L
Let Ti be the inside temperature, T1 be the temperature at the plasterboardbrick interface, T2
be the temperature at the brickwood interface, and To be the outside temperature. Then
kpTi − kpT1 = kbT1 − kbT2 (1) kbT1 − kbT2 = kwT2 − kwTo (2) and Solving (1) for T2 gives
T2 = (kp + kb)T1/kb − (kp/kb)Ti
a. Substituting this into (2) and solving for T1 yields
T1 = ( kp /kb ) (1 + kw /kb ) Ti + ( kw /kb ) T0 =
(1 + kw /kb ) (1 + kp /kb ) − 1 21 °C b. Using this value in (1) yields T2 = 18 °C 14. REASONING If m kilograms of ice melt in t seconds, then Q = mLf (Equation 12.5) joules
of heat must be delivered to the ice through the copper rod in...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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