Solutions to Homework Set 13 Webassign Physics 105

A venturi tube is a tube with a constriction in it. Pressure in a venturi tube can be measured by attaching a U-shaped fluid-filled device to the venturi tube as shown in the figure above.The density of air is 1.20 kg/m3 and the density of water is 1000 kg/m3. Use g = 10 m/s2. (a) If the fluid in the U is water, and there is a 10.0 cm difference between the water levels on the two sides, what is the magnitude of the pressure difference between points 1 and 2 in the venturi tube?

The pressure at the bottom dotted line must be equal in the two tubes. Therefore the pressure pushing down non the left must equal the pressure pushing down on the right plus the height of the water on the right above the bottom dotted line.

Here this equals 1000 Pascals

(b) The venturi tube has air flowing through it. If the cross-sectional area of the venturi tube is 3 times larger at point 1 than it is at point 2, what is the air speed at point 2?

First apply Bernoulli's equation

Since the points of interest are at the same height, h1 = h2 and can be removed from the equation. In addition, by rearranging to form P1-P2 and inserting the solution to part (a) v2 can then be solved.

2) While taking a shower, you notice that the shower head is made up of 42 small round openings, each with a radius of 2.00 mm. You also determine that it takes 2.00 s for the shower to completely fill a 1.00-liter container you hold in the water stream. The water for the shower is pumped by a pump that is 5.10 m below the level of the shower head. The pump maintains an absolute pressure of 1.50 atm. Use g = 10 m/s2, and assume that 1 atmosphere is 1.0 x 105Pa. (a) At what speed does the water emerge from the shower head?

For this problem, apply the flow rate equation: V1A1 = V2A2 = volume per time Taking the given time for 1 L of water to flow into the container as t = 2 sec, V1A1 = 1L/t. Thus the velocity per hole times the total area of all the holes must equal 1L/t.

(b) What is the speed of the water in the pipe connected to the pump?

Applying Bernoulli's equation with given h1, h2, P1, P2 and v2, the solution for the velocity at the pump is:

Applying the flow rate equation, A1 = V2A2/V1

3) A ring made from iron has an inner radius of 2.50000 cm and an outer radius of 3.50000 cm, giving the ring a thickness of 1.00000 cm. The thermal expansion coefficient of iron is 12.0 x 10-6/°C. If the temperature of the ring is increased from 20.0°C to 75.0°C, by how much does the thickness of the ring change?

The thickness of the ring is determined by the difference between the outer radius and the inner radius. In general the new length is the original length times the change in temperature times the coefficient of expansion. Thus:

For this problem, T = 1+ 12E-6 * 55 * 1 = 1.00066 cm.

4) A copper block with a mass of 600 grams is cooled to 77K by being immersed in liquid nitrogen. The block is then placed in a Styrofoam cup containing some water that is initially at +50.0°C. Assume no heat is transferred to the cup or the surroundings. The specific heat of liquid water is 4186 J/(kg °C), of solid water is 2060 J/(kg °C), and of copper is 385 J/(kg °C). The latent heat of fusion of water is 3.35 x 105 J/kg. When the system comes to equilibrium, the temperature is 0°C.

(a) Calculate the maximum possible mass of water that could be in the cup.

Two solutions to this problem exist. The first, that has the maximum possible water in the cup, is found by saying that just enough water is present such that the copper block brings the water from an initial temperature of 50C to a final temperature of 0C. (b) Calculate the minimum possible mass of water that could be in the cup. grams

5) 500 g of water at 20°C is in a pot on the stove. An unknown mass of ice that is originally at -10°C is placed in an identical pot on the stove. Heat is then added to the two samples of water at precisely the same constant rate. Assume that this heat is transferred immediately to the ice or water (in other words, neglect the increase in temperature for the pot). We will also neglect evaporation. The ice melts and becomes water, and you observe that both samples of water reach 50.0°C at the same time. (a) How does the mass of the original piece of ice in the second pot compare to the mass of the water in the first pot?

Since the total heat added is the same for each situation, writing the heat transfer equation for each situation is extremely instructive:

For Water on the stove:

For Ice on the stove:

Since the last three terms in the equation for the ice each contain a factor of mass, the mass of the water must be less than the mass of the water in the first equation.

CORRECT The mass of the ice is larger than the mass of the water. INCORRECT The mass of the ice is the same as the mass of the water. INCORRECT The mass of the ice is smaller than the mass of the water. INCORRECT There's not enough information to answer this question.

(b) Which system will reach 80°C first?

Again, since the mass of the now melted ice is less than the mass of the water, it is able to heat from 50C to 80C with less consumed heat and therefore in less time.

INCORRECT The first pot, which had the 20°C water in it at the beginning. CORRECT The second pot, which had the ice in it at the beginning. INCORRECT When the pots reach 50.0°C, they both contain water, so they will both reach 80°C at the same time. INCORRECT There's not enough information to answer this question.

(c) Solve for the mass of the ice that was originally in the second pot. The specific heat of liquid water is 4186 J/(kg °C), and of solid water is 2060 J/(kg °C). The latent heat of fusion of water is 3.35 x 105 J/kg.

Using the equations formed for part (a), and substituting the heat from the stove in the second equation with the heat from the stove in the first equation yields: Where "water1" is the water that begins at 20C and "water2" is the water created by melting the ice.

6) In 1992, a Danish study concluded that a standard toy balloon, made from latex and filled with helium, could rise to 10000 m (where the pressure is 1/3 of that at sea level) in the atmosphere before bursting. In the study, a number of balloons were filled with helium, and then placed in a chamber maintained at -20°C. The pressure in the chamber was gradually reduced until the balloons exploded, and then the researchers determined the height above sea level corresponding to that pressure. Assume each balloon was filled with helium at +20°C and at about atmospheric pressure. Determine the balloon's volume just before it exploded, if its volume when it was first filled was 520 cm3.

Here, there are two states of the balloon. The initial state is at +20C, 1 atm and has a given volume. The second is at -20C, 1/3 atm and the volume is the variable of interest. Thus, since the number of particles in the balloon remains constant:

solving for V2 yields:

or for the general case of this problem:

3 V2 = 2.6 V1 = 1347 cm

7) Warning: the volume shown by the simulation is incorrect. To fix it, multiply the volume given in the simulation by a factor of 10. The cylinder contains a certain number of moles of monatomic ideal gas in a particular initial state. If you add 120. J of heat at constant volume, the temperature increases. Starting from the same initial state, how much heat should be added at constant pressure to achieve the same final temperature? Use principles of physics to determine the answer and then confirm it using the simulation.

To do this problem without the simulation it should be noted that the ideal gas law applies. For the first process, since no work is done, the internal energy is changed by adding in heat directly. That is, the change in U = Q.

So,

For the second situation which yields

by plugging in the known Q from the first equation:

Or here 200 J.

8) A system of ideal gas has an initial pressure of 115 kPa and occupies a volume of 2.00 liters. Doubling the system’s absolute temperature by means of a constant-pressure process would require an amount of work W. Instead, you decide to double the absolute temperature by carrying out two processes in sequence, a constant-pressure process followed by a constant-volume process. In this case, the total work done in the two-process sequence is W/2. Find the final pressure of the system.

The key equations to this question are the ideal gas law and the work done by a constant pressure process:

Thus, for the first process, being constant pressure, the final volume must be twice the initial volume since the temperature doubles. This dictates that the change in volume must be 2V-V = V.

For the second situation, if half the work is done, then the change in volume may only be half of the previous change in volume. That is, a change in volume of V/2. This gives a final volume of 3V/2 or an increase in volume of one and a half. Thus, by the ideal gas law, if the volume increases by 3/2 and the temperature doubles, the change in pressure must be 4/3. 4/3 x 3/2 = 2.

Thus, for this problem the final pressure is 4P/3 = 153 1/3 Pa.

(b)Find the final volume of the system.

As stated in part (a) the volume increases by a factor of 3/2. Thus the new volume is 3V/2 = 3 L.

9) Consider the four-process cycle shown in the P-V diagram in the figure above. The graph shows a sequence of four processes being carried out on a sealed system of ideal gas. In this case, P is 50.0 kPa and V is 1.00 liters. (a) Calculate the work done by the gas in the process taking the system from state 1 to state 2.

The work done in taking the system from state 1 to 2 is defined by the area beneath the line that connects those two points. In this case it is the area of a trapezoid:

or here, 2P x V = 100 J

(b) Calculate the work done by the gas in the process taking the system from state 2 to state 3.

Again the work done by the gas is equal to the area underneath the line. Here it is a constant pressure process and therefore:

or here, 3P x V = 150 J

Since this is a constant volume process, there is no area under the curve and hence no work.

(d) Calculate the work done by the gas in the process taking the system from state 4 to state 1.

As before, the work is given by the area of a trapezoid but, because the process is proceeding from right to left it is negative work. or here 3P/2 x 2V = 150 J

(e) Calculate the net work done by the gas in one entire cycle.

The net work is W1-2 + W2-3 + W3-4 + W4-1 = 100 + 0 + 150 - 150 = 100 J

(f) Calculate the net change in the internal energy of the gas for one entire cycle.

The system returns to its starting point and thus the change in temperature is zero. Therefore the heat added to the system is 0. By following the first law of thermodynamics:

Thus the change in internal energy = -100 J

(g) Calculate the net heat added to the gas for one entire cycle.

The system returns to its starting point and thus the change in temperature is zero. Therefore the heat added to the system is 0

(h) Complete this sentence - The temperature in state 3 is larger than the temperature in state 1 by a factor of .

Between states 1 and 3 the pressure has tripled and the volume has tripled. Thus the temperature must have risen by a factor of 9.

For the last two parts, assume that the ideal gas in this system is monatomic. (i) Calculate the change in internal energy experienced by the gas during the process that takes the system from state 1 to state 2.

For an ideal, monatomic gas, the internal energy is equal to 3/2nRT. Thus the change in temperature is given by 3/2nR(Tf-Ti) or 3/2(PfVf-PiVf) or here 3/2(6PV-PV) = 15PV/2 = 375 J

(j) Calculate the heat added to the gas for the process that takes the system from state 1 to state 2.

By the first Law of Thermodynamics, the heat is the change in internal energy plus the work. This is 375 + 100 = 475 J.