TECHNICAL PHYSICS II TEST NO. 1 SPRING 2008
1. Equal amounts of ice at 0 deg C and water at 80 deg C are placed
in an insulated container. What is the temperature when the system
comes into thermal equilibrium? Is any ice left? If so, what
fraction of the original ice is left? The heat of fusion for water
is 79.6 cal/g and the specific heat is 1.00 cal/g-K.
You have to check the two possibilities. Either the ice all melts or
it doesn't. One way to approach the problem is to see how much heat
it takes to melt the ice and then see if that much energy can be
realized by the cooling of the water down to 0° C.
Heat needed to melt ice is:
Q = m(79.6 cal/g),
where m is the (unknown) amount of ice.
Heat given off by cooling of water to 0° C:
Q = m(1 cal/g·° C)(80° C) = m(80 cal/g).
Hence all the ice will melt. To find the final temperature note that
by conservation of energy, the amount of heat needed to melt the ice
plus raise the meltwater to the final temperature, T, equals the heat
lost by the hot water. Hence,
m(79.6 cal/g) + m(1 cal/g·° C)(T - 0° C) = m(1 cal/g·° C)(80° C - T).
The m's cancel out. Solving for T gives 0.200° C. (This problem
was taken from another book on faith. I didn't realize the numbers
would come out a little screwy.)
2. A wall is made up of a 10.0-cm layer of red brick (thermal conduc-
tivity k = 0.630 W/m-K) in contact with a 20-cm layer of cinder block
concrete (k = 0.700 W/m-K). If the inside concrete face has a tempe-
rature of 20 deg C and the outside brick face is at 0 deg C, what is
the heat current per unit area through the wall?
The heat current must be the same for the brick and the cinder block.
Hence,
(0.630 W/m·K)A(T - 0° C)/(0.10 m) = (0.700 W/m·K)A(20° C - T)/(0.20 m),
where T is the interface temperature. We need T to find the heat
current. The area A cancels out. Solving,
T = 7.14° C.
Calculate the heat current per unit area.
H/A = (0.630 W/m·K)(7.14° C)/(0.100 m) = 45.0 W/m2.
Check:
H/A = (0.700 W/m·K)(20° C - 7.14° C)/(0.200 m) = 45.0 W/m2.
3. Consider a cyclic process on a P vs V diagram: The system's pressure
is reduced at constant volume from state a to state b. Then the
pressure is held constant as the system contracts from b to c. The
volume is again held constant and the pressure is increased until
it attains its original value at state d. Finally, the pressure is
held constant as the system expands back to state a. The pressure
and volume at state a are 2.50E+6 Pa and 0.00500 cubic meters, respec-
tively. The pressure and volume at state c are 5.00E+5 Pa and 0.00250
cubic meters, respectively. Along paths a-b, b-c, and c-d the heat
flow was -3000 J, -4000 J, and +5500 J, respectively. What was the
internal energy change of the system along each of the four paths and
the heat flow along path d-a? Was this flow into or out of the system?
(It would help to sketch a P versus V diagram.)
For a → b, you are at constant volume, hence W = 0. From the
first law of thermodynamics,
Q (= -3000 J) = ΔU + W = ΔU: ΔU = -3000 J.
For b → c, you do a negative amount of work. Since the pressure
is constant the work is
W = PΔV = (5 × 105 Pa)(0.0025 m3 - 0.0050 m3) = -1250 J.
An amount of heat (-4000 J) was transferred. By the first law,
-4000 J = ΔU + -1250 J: ΔU = -2750 J.
For c → d, you are again at constant volume. From the first law,
5500 J = ΔU
For d → a, you have to have a change in internal energy such that
∑ΔU = 0 for the cycle. This means
ΔU = 250 J.
Work is done from d to a, given by PΔV. Applying the first law,
Q = 250 J + (2.5 × 106 Pa)(0.005 m3 - 0.0025 m3) = 6500 J.
Since the result is positive, the heat flows into the system.
4. Four identical charges, each with a charge of 15.0 microcoulombs, are
located on the corners of a rectangle at (x,y) positions given by
(0,0), (0, 0.150 m), (0.350 m, 0), and (0.350 m, 0.150 m). Calculate
the magnitude and direction of the resultant electric field at the
origin (0,0).
The problem should have read, "Calculate the magnitude and direction
of the resultant electric field at the origin due to the three other
charges and the force on the charge at the origin." If you calculated
the electric field at the origin due to the three distant charges,
that counts as a solution. For a point charge, of course, the
electric field is "infinite" (undefined) at it's location.
Number the charges 0 to 3, with 0 being the number of the charge at
the origin. The forces are then due to charges 1, 2, and 3 on charge
0. Call the fields E1, E2, and E3.
E1 is directed in the negative y direction, E2
is directed into the third quadrant, and E3 is in the
negative x direction. The magnitudes of the fields are given by the
Coulomb formula
E = kq / r2.
The distance between the charge at the origin and each of the other
three charges is 0.15 m, [(0.15 m)2 + (0.35)2]1/2 = 0.381 m,
and 0.35 m, respectively. Putting the numbers in the Coulomb formula
you get
E1 = 6.00 × 106 N/C
E2 = 9.31 × 105 N/C
E3 = 1.10 × 106 N/C
You have to add the fields vectorially. This means
E = [(∑Ex)2 + (∑Ey)2]1/2,
to get the magnitude, which requires you to find the x and y
components of each force. The direction is given by
θ = tan-1|∑Ey / ∑Ex|,
with respect to the x axis in the quadrant containing the resultant
field. The only field not along an axis is E2. The angle
between the direction of this force and the negative x axis is given
by
θ = tan-1(0.15 m / 0.35 m) = 23.2°.
Hence,
E2x = -(9.31 × 105 N/C)cos(23.2°) = -8.86 × 105 N/C
E2y = -(9.31 × 105 N/C)sin(23.2°) = -3.67 × 105 N/C
Now you can sum the components.
∑Ex = -1.10 × 106 - 8.86 × 105 = -1.99 × 106 N/C
∑Ey = -6.00 × 106 - 3.67 × 105 = -6.37 × 106 N/C
The equations above give
E = 6.67 × 106 N/C at 73° below the -x axis.
5. A spherically-symmetric charge distribution consists of an insulating
sphere 1.00 cm in radius containing a charge of +1.00 nC distributed
uniformly throughout its volume, surrounded concentrically by an
insulating spherical shell with an inner radius of 1.00 cm (such that
the sphere and shell are in contact) and an outer radius of 2.00 cm
containing a charge of -2.00 nC distributed uniformly throughout its
volume. What is the magnitude and direction of the electric field
at a distance r = 1.50 cm from the center of this charge distribution?
Recall that the volume of a sphere is (4/3)*pi*r^3.
Gauss' law shows that a spherical distribution of charge looks like
a point charge if you are outside that distribution. You have to
calculate how much charge is inside the r = 1.50 cm radius and then
use the point charge formula.
The sphere is entirely inside r = 1.50 cm, so all its charge, +1.00 nC,
contributes. The volume of the shell inside r = 1.50 cm is equal to
ΔV = V1.5 - V1.0
where V1.5 is the volume of a sphere of radius 1.50 cm and
V1.0 is the volume of a sphere of radius 1.00 cm. The
fraction of the volume inside r = 1.50 cm is
(V1.5 - V1.0) / (V2.0 - V1.0),
where V2.0 is the volume of a sphere of radius 2.00 cm.
This evaluates, due to the formula for the volume of a sphere, as
[(1.50 cm)3 - (1.00 cm)3] / [(2.00 cm)3- (1.00 cm)3] = 0.339.
Since the charge density is uniform, the amount of charge in the shell
inside r = 1.50 cm is
q = (-2.00 nC)(0.339) = -0.679 nC.
The total charge inside r = 1.50 cm is therefore 1.00 nC - 0.679 nC
= 0.32 nC. The electric field points outward due to the net positive
charge. Its magnitude is
E = (9 × 109 N·m2/C2)(0.32 × 10-9 C) / (0.015 m)2 = 1.28 × 104 N/C