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course Phy 202
submitted 03/04
Ph2 Class 12130
If you haven’t done so for another course, work through and submit the following documents, which review the calculation
of areas and volumes. Answer and self-critique the questions you aren’t sure of; you can just OK those you are sure of:
areas
volumes
`q000. Report the data you obtained in today’s class, where we measured temperature vs. clock time for initially warm
water in a cup. After some time elapsed we introduced some ice into the cup, and we later introduced a very warm steel
washer, continuing to measure the temperature as a function of time. After providing your raw data, detail how you then
used your data to calculate the specific heat steel. Include estimates of uncertainty.
warm water
temp (degree Celsius) time (seconds)
20 0
30 10
34 20
35 30
35 60
34 120
33 180
33 240
32 300
32 360
small ice cube 30 370
large ice cube 24 400
21 430
20 460
21 490
hot washer 23 500
24 550
23 610
Hot water with with a room temperature washer dropped into the cup
temp (degrees Celsius) time (seconds)
89 0
75 30
75 90
75 150
73 210
hot water control with no room temperature washer dropped into the cup
temp (degrees Celsius) time (seconds)
89 0
84 30
80 90
77 150
`q001: What is the weight of a column of water 10 cm high, in a vertical tube of diameter 3 mm?
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volume is pi r^3 h
so, pi (1.5 mm)^3(100 mm)
=337.5 mm^3
998.2 kg/m^3 density of water at 20 degrees Celsius,
So, 0.0000003375 mm^3 x 998.2 kg/m^3
=0.0003369 kg
then mass times gravity will give you weight
So, 0.0003369 kg * 9.8 m/s^2=0.003302 N is the weight of the water.
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The volume of water in the tube is (cross-sectional area) * length.
The cross-sectional area is pi r^2, not pi r^3.
Your calculation pi r^3 h would give you units of mm^4, not mm^3.
You also failed to multiply by pi.
Your reasoning is fine; with those two corrections you'll be good up to this point.
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#$&*
How much force is required to support that column?
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0.003302 N * .1 m = 0.0003302 Joules
@&
To support the water you need a force equal to its weight.
N * m = Joules would be a unit of work, not relevant to this calculation.
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#$&*
If the column is supported by water at the base of the tube, what must be the pressure at the base?
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pressure is gravity times density times height of tube
So, approx 10 m/s^2 for gravity and 1000 kg/m^3 for density and 0.1 m for altitude
10m/s^2 (1000 kg/m^3)(.1 m) = 1000 N/m^2
@&
Right, but the result can be obtained from your previous results (if corrected)
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What pressure would be required to support a column 50 cm high?
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10 m/s^2 (1000 kg/m^3)(.5 m) = 5000 kg m/s^2
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Symbolic alternative:
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Let rho stand for the mass density of water and g for the acceleration of gravity.
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A column of water in a vertical cylindrical tube has height y. The tube has radius r.
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How much force is required to support the water in the tube?
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rho = mass density of water
g = gravity
y = height of column
r = radius of tube
first is volume with, V = pi r^3 y
then mass by m = V*rho
then weight by W = m * g
then force by F = W * h
@&
the weigth is a force; if you multiply it by the height you get work or energy which, however, does not apply to this situation.
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#$&*
Using your preceding answer, what is the pressure at the base of the tube?
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rho * g * y = pressure at the base of the tube.
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`q002. Water exits a cylindrical container, whose diameter is 6 cm, through a hole whose diameter is 0.3 cm. The speed
of the exiting water is 1.3 meters / second.
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How long would it take the outflowing water to fill a tube having cross-sectional diameter 0.3 cm and length 50 cm?
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for the tube being filled
pi r^2 h
is pi (.15 cm)^2 (50 cm)
1.125 cm^3 is the volume
@&
this doesn't answer the question of how long it would take to fill the tube
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What volume of water exits during this time?
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the tube it filled has a radius of 0.15 and and length of 50 cm so,
V = pi (.15 cm)^2(50 cm)
V = 1.125 cm^3
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By how much does the water level in the cylinder therefore change?
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not too much but it would be noticeable to the eye.
@&
you know the diameter of the cylinder so you can calculate the change in depth.
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What is the ratio of the exit speed of the water to the speed of descent of the water surface in the container?
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@&
you need the preceding result to answer this
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Symbolic alternative:
Water flows from a cylinder whose radius is R_cyl through a hole of radius r_hole, with water level y relative to the hole.
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A short tube of length `dL and diameter equal to that of the outflow hole extends horizontally from the hole.
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What mass of water is required to fill the tube?
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Volume = pi r_hole ^2 * dL
then, mass = V * rho
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What is the PE change of the system during the interval required to fill the tube?
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W = mass * gravity
@&
That's the weight of the water, not its change in PE.
Hint: Previously you did calculations simiilar to the one you need here, but they weren't appropriate in answer to those questions.
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#$&*
Based on these results, assuming no dissipative forces to be present, what is the velocity of the water flowing from the
cylinder?
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Weight * y = joules of pressure (energy)(KE or PE)
velocity = sqrt (2 * KE / m)
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How long does it therefore take for the cylinder to fill?
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velocity / cubed rt of Volume = time to take to fill the cylinder.
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What is the expression for the volume rate at which water flows from the cylinder, with respect to time?
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What is the expression for the rate `dy / `dt at which the depth of water in the cylinder changes with respect to time?
`q003. Assume the water level in the container of the preceding question is 50 cm above the outflow hole.
What mass of water is required to fill the 50 cm tube?
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Volume = pi (.15 cm)^2(50 cm)
V= 3.53 cm^3
m = 0.0000353 m^3 (1.00 kg/m^3)
0.000353 kg
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As that mass of water exits the cylinder, what is the change in the gravitational PE of the system?
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PE = 0.000353 kg * 9.8 m/s^2
PE = 0.003459 Joules
@&
This is how you calculate the weight of the water, not its change in PE.
kg m/s^2 is not Joules
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#$&*
What therefore is the speed of the exiting water?
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sqrt (2*0.003459 J / 0.000353 kg) = 4.427 m/s
@&
if you get the PE change right this will work
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How long does it therefore take to fill the 50 cm tube?
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By what percent will the water level in the cylinder change during this time?
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At what rate is water flowing from the cylinder, in units of cm^3 / second?
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Symbolic alternative:
If you have completed the symbolic alternative to the preceding problem, you may use your results along with the given
quantities to answer this question.
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`q004. In a cylinder containing a fixed large number of particles, all colliding elastically with one another and with
the container, one end of the cylinder consists of a moveable piston.
If the piston is stationary, then does the speed of a particle increase, decrease or remain the same when it collides
with the piston?
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it should remain the same becuase the piston is not moving.
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If the piston is moved towards the other end of the cylinder:
Does the speed of a particle increase, decrease or remain the same when it collides with the piston? (and do you see a
potential discrepancy between what you know happens and the previously given definition of an elastic collision?)
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particle speed would increase because the volume of the cylinder would decrease as the piston moves to the other side.
@&
particle velocity increases because it collides elastically with a large approaching mass, not because volume decreases
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#$&*
Does this result in more pressure, no change in pressure or less pressure on the piston?
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This would result in more pressure on the piston
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Does the volume of the system increase or decrease in this process?
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The volume of the system would decrease as the piston closes in on the opposite side.
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`q005. The gas in a system is confined if no gas can enter or leave the system. So the number of particles in a
confined gas does not change.
If the pressure in a confined gas is held constant while the temperature is increased what happens to its volume?
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increasing temperature will increase the volume of a gas
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If the volume of a confined gas is held constant while the temperature is increased what happens to its pressure?
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pressure will increase when the temperature is increased.
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If the temperature in a confined gas is held constant while the volume is increased what happens to its pressure?
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When the volume is increased the pressure will decrease.
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`q006. For a confined gas, a temperature change can result in pressure changes, volume changes, or both pressure and
volume changes.
If the percent changes in the pressure, volume and temperature of a confined gas are small, then every 1% increase in
pressure implies a 1% contribution to an increase in absolute temperature, while every 1% increase in volume implies a 1%
contribution to a decrease in temperature. So for example a 1% increase in pressure accompanied by a 1% decrease in
volume would imply a 2% increase in temperature, while a 1% decrease in pressure accompanied by a 1% decrease in volume
would imply unchanged temperature.
If volume increases by 2% and pressure increases by 3%, what is the percent increase in temperature?
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since volume and pressure and inversely related we need to subtract the volume from the pressure so,
3% - 2% = 1% increase in temperature.
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If volume decreases by 2% and pressure increases by 3%, what is the percent increase in temperature?
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again they are inversely related so we need to add the volume to the pressure.
so,
2% + 3% = 5% increase in temperature
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What must happen to the volume when combined with a 3% increase in temperature, if the result is to be a 1% decrease in
pressure?
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need to do some algebra here so
3% temp inc + volume = 1% dec pressure
volume needs to increase 4% to decrease pressure by 1%
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What must happen to the pressure when combined with a 3% increase in temperature, if the result is to be a 2% decrease in
volume?
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solve algebraically so
3% inc temp + pressure = 2% dec volume
Pressure needs to be 5% increase.
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`q007. On a set of x and y axes:
Sketch dots at (-5, 1), (0, 1) and (2, 0). Connect the dots from left to right with straight line segments, and when you
get to (2, 0) keep going for a ways. Call this ‘path 1’.
Sketch a straight line from (-5, 1) to (0, 0) and keep going until the line intersects path 1, and keep going for a
little ways. If you need to extend path 1 a bit in order for the lines to intersect, do so. Call your new path ‘path
2’.
At what coordinates do the paths intersect?
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(3, -1)
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Sketch a dot at (-2, 0). Sketch a line from (-5, 1) through this dot and continue until the line hits the y axis. Make
a dot at this point. Then sketch a horizontal line starting at this last dot, directed to the right and continue the
line until it intersects each of your other two paths.
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What are the coordinates of your points of intersection?
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(3, -1)
@&
The actual point wouldn't be (3, -1), but (3, -1) is a good estimate, which is all that's required here.
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Symbolic alternative:
Path 1 is defined as y = a if x < 0, y = a - b x if x > 0, where a and b are positive constants.
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Path 2 is the line y = m x, where m is a negative constant.
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For x < 0 path 3 is the line through (-a/b, 0) having slope m, where -b < m < 0. The path is continuous for all x, and
for x > 0 the path is parallel to the x axis. At what points does this path cross each of the first two paths?
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`q008. Two identically-shaped cylinders, each of diameter 2 centimeters, hang straight down into the water, extending 5
centimeters beneath the surface. Water pressure results in forces acting perpendicular to the surfaces of each cylinder.
When all these forces on a cylinder are added up, the result is a buoyant force on that cylinder.
One cylinder is made of steel, the other of glass.
Does the pressure exerted at a point on the lower end of the cylinder depend on whether the material is steel or glass?
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yes because each has a different density and the weight of each material is acting against the force of the water holding the material afloat
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the water doesn't know what material it's supporting; the pressure depends only on water density and depth. These quantities are the same for both materials.
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Does the pressure exerted on the side of the cylinder, at a given depth, depend on whether the material is steel or
glass?
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No the pressure on the side will not determine if the material will sink or float.
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Does the buoyant force on the cylinder depend on whether the cylinder is steel or glass?
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Yes steel and glass has different densities.
@&
the water doesn't know what material it's supporting; the pressure depends only on water density and depth, and the force depends only on area. These quantities are the same for both materials.
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#$&*
If the glass or steel cylinder is removed, the surrounding water would quickly occupy the vacated cylindrical region. So
in that sense we would now have a cylinder of water in place of the glass or steel cylinder. The ‘water cylinder’
experiences a buoyant force from the pressure of the water surrounding it, just as the glass or steel cylinder
experienced a buoyant force. Is the buoyant force on the ‘water cylinder’ greater or less than that on the glass and
steel cylinders?
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The buoyant force on the water cylinder would be the same as the force exerted on the glass and steel cylinders. Its the
weight and densities of glass, steel and water cylinders that makes it different.
@&
Correct. However the weight of the steel or glass has no effect on the buoyant force.
They would have an effect on the net force
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#$&*
The glass and steel cylinders would have sunk of the buoyant force was the only force acting on them. Something in
addition to the buoyant was supporting them either from above or from below. However no additional force is required to
support the ‘water cylinder’. How therefore does the buoyant force of the ‘water cylinder’ compare to the weight of the
water in that cylinder?
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the buoyant force against the water cylinder would be stronger than the weight of the water cylinder.
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How does the buoyant force on the glass or steel cylinder compare to the weight of the cylinder?
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They would be significantly more because they have a much higher density.
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How does the buoyant force on the glass or steel cylinder compare to the weight of the water displaced by the cylinder?
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the weight of the glass and steel cylinders are more than the buoyant forces of the water displaced.
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@&
The buoyant force in every case would be identical.
We conclude that the buoyant force is equal to the weight of the displaced water.
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University Physics: A sphere of radius a, centered at the origin, is submerged in water, whose density will be
symbolized as rho_water. The water surface is the plane z = h, where h > a. For any small region of the sphere’s
surface, the water pressure exerts a force perpendicular to that region equal to P_ave * `dS, where `dS is the area of
the region and P_ave is the average pressure at the depth of that region.
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For a small region `dS containing point (x_i*, y_j*, z_k*), with outward unit normal N:
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What is the pressure at the point?
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Assuming that the average pressure does not differ significantly from the pressure at the given point, and also that the
region is small enough that it can be considered flat, what is the magnitude of the force on the region? What is the
vector force on the region?
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What therefore is the total force exerted by pressure on the sphere? Hint: Start by partitioning the sphere, in
whatever coordinate system you choose. The bounds on the integral are simplest in spherical coordinates, most
complicated in rectangular; the symmetries involved might be simplest in cylindrical coordinates. However since you are
probably more familiar with resolving vectors in the rectangular coordinate system, that will probably tip the scales in
favor of rectangular coordinates.
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`q009. A 400 gram rock at 100 Celsius is dropped into 2 liters of water at 25 Celsius. The temperature of the
surrounding air is 30 Celsius.
After about 20 minutes the temperature of the system reaches a maximum of 40 Celsius. If we assume that during this time
the system doesn’t exchange a significant amount of energy to the surrounding air, what do we conclude is the average
specific heat of the material in the rock (e.g., how many Joules would it take to raise the temperature of one gram of
rock by 1 Celsius degree)?
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400 gram rock is .4 kg
heat added was 4.18 J/g Celsius * 2000 grams * 15 Celsius = 125400 joules
c = specific heat
125400 J = 400 grams * c* 60 Celsuis
c = 5.225 J/g C for the rock
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At 25 Celsius the temperature of the 2 liters of water was rising, in response to the greater temperature of the air, at
a rate of .01 degree Celsius per minute. Making reasonable assumptions and approximations, attempt to adjust your result
for the specific heat accordingly. The limit on the amount of insight it is possible to bring to this situation probably
exceeds the amount of time you can reasonably spend on it. Overlooking the fact that insight and time and not
commeasurable quantities, balance the time you spend on this question accordingly.
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It would have taken a very short time in a matter of seconds to increase the temperature of the water above the room
temperature 30 Celsius. The one hundredth of a degree increase per minute would not have a big effect bc the system
would be above the surrounding air temperature in less than a minute.
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Symbolic alternative:
Mass m_1 of a material at temperature T_1_0 is immersed in mass m_2 of another material, whose specific heat is c_2 and
whose initial temperature is T_2_0. The system is insulated from its surroundings, and comes to final temperature
T_final. What is the specific heat of the first material?
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c_2 * m_2 * (T_2_0 - T_final) = added heat
added heat/ (m_1 * (T_1_0 - T_final) = specific heat of material
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Pretty challenging at the level of University Physics: Suppose that the ambient temperature is T_room and the rate of
thermal energy exchange between the system and the room, when the water is at temperature T, is dQ_2/dt = k_2 e^(T -
T_room). Suppose also that the rate at which the first sample exchanges thermal energy with the second is dQ_1/dt = k_1
e^(T_2 - T_1), where T_2 and T_1 are the (variable) temperatures of the two samples. For the situation of the original
problem, what are reasonable values of k_1 and k_2?
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`q010. A long thin plastic cylinder with walls whose thickness is 5% of the cylinder’s radius is filled with water at
100 Celsius and immersed in an ice water bath. The temperature of the water in the cylinder is observed to decrease by
10 Celsius in 30 seconds.
The experiment is to be repeated with a cylinder of the same dimensions as the first, whose walls are twice as thick as
the first. About how much temperature change would you expect in 30 seconds?
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since the cylinder wall is twice as thick it should take twice as long to cool off so the water would decrease by 5
Celsius in 30 seconds.
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The original plastic cylinder is replaced by a glass cylinder whose dimensions are the first, and the temperature of the
water is observed to change by 10 Celsius in 10 seconds. Do we conclude that glass conducts thermal energy better than
plastic, or plastic better than glass?
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The glass conducts thermal energy better than plastic.
@&
You need to give a basis for this.
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If the experiment was repeated with a glass cylinder whose dimensions are the same as that of the second plastic
cylinder, how long do you think it would take the temperature of the water to change by 10 Celsius?
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I think it would double so it would take 20 seconds to decrease the temperature by 10 Celsuis.
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How would the rate at which the temperature of the 100 C water inside the cylinder changes be related to the thickness of
the walls?
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Directly related to the thickness of the wall...the thicker the wall the longer it will initially take to see a change in
temperature of the water.
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University Physics Students:
In what ways might the ratio thickness of the cylinder’s walls to the radius of the cylinder affect your answers to the
preceding question?
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@&
Pretty good overall, but you should make some revisions. I've inserted a number of notes.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
Spend a reasonable amount of time on your revision, but don't let yourself get too bogged down. After a reasonable amount of time, if you don't have at least a reasonable attempt at a solution, insert the best questions you can showing me what you do and do not understand, and I'll attempt to clarify further. &#
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