PHYSICS COLLEGE

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer 1

Answer:

Answer:149.73 ml

Explanation:

Given

$\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)$

change in volume is given by

$\Delta V=V_(final)-V_(initial)$

$\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]$

$V_(final)=149.73 ml$

Answer 2

Answer:

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

Learn more about Volume calculation here:

brainly.com/question/33318354

#SPJ3

Related Questions

A 750-kg automobile is moving at 26.2 m/s at a height of 5.00 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?
Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....? Listed following are locations and times atwhich different phases of the Moon are visible from Earth’sNorthern Hemisphere. Match these to the appropriate moon phase. The three given moon phases are: Waxing Crescent Moon Waning Crescent Moon Full Moon The things we need to match to the above three topicsare: *visible near western horizon an hour after sunset *rises about the same time the sun sets *visible near eastern horizon just before sun rises *occurs about 3 days before new moon (i know this is waningcrescent) *visible due south at midnite *occurs 14 days after new moon (i know this is full moon)
An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?
Suppose that you run along three different paths from location A tolocation B. Along which path(s) would your distance traveled be different than your displacement (Question 2) please help
A rigid tank initially contains 3kg of carbon dioxide (CO2) at a pressure of 3bar.The tank is connected by a valve to a frictionless piston-cylinder assembly located vertically above, initially containing 0.5 m^3 of CO2. The piston area is 0.1 m^2. Initially the pressure of the CO2 in the piston-cylinder assembly is 2 bar. The ambient pressure and temperature are 1 bar and 290 K. Although the valveis closed, a small leak allows CO2 to flow slowly into the cylinder from the tank. Owing to heat transfer, the temperature of the CO2 throughout the tank and the piston-cylinder assembly stays constant at 290K. You can assume ideal gas behavior for CO2.Determine the following:a. The total amount of energy transfer by work (kJ)b. The total amount of energy transfer by heat (kJ)

If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?

Answers

Answer:

$heigth=2.86m$

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

$V_(avg)=(distance)/(time)\nV_(avg)=(1.6m)/(0.19s)\nV_(avg)=8.42m/s$

Also we know that average velocity

$V_(avg)=(V_(i)+V_(f))/2\n$

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

$V_(f)=V_(i)+gt\nV_(f)=V_(i)+(9.8)(0.19)\nV_(f)=V_(i)+1.862$

Substitute value of Vf in average velocity

$V_(avg)=(V_(i)+V_(f))/2\nwhere\nV_(f)=V_(i)+1.862\nand\nV_(avg)=8.42m/s\nSo\n8.42m/s=(V_(i)+V_(i)+1.862)/2\n2V_(i)+1.862=16.84\nV_(i)=(16.84-1.862)/2\nV_(i)=7.489m/s\n$

Vi is speed of balloon at top of the window

Now we need to find time

$V_(i)=gt\nt=V_(i)/g\nt=7.489/9.8\nt=0.764s$

So the distance can be found as

$distance=(1/2)gt^(2)\n distance=(1/2)(9.8)(0.764)^(2)\n distance=2.86m$

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

$f = f_0 ((v_0)/(v_0-v))$

Here,

$f_0$ = Frequency of Source

$v_s$ = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

$f = f_0 ((v_0)/(v_0-v))$

$300 = f_0 ((343)/(343-v))$

$(300*343) - 300v = 343f_0$

Now the second expression will be,

$f' = f_0 ((v_0)/(v_0-v/2))$

$290 = (343)((v_0)/(343-v/2))$

$290*343-145v = 343f_0$

Dividing the two expression we have,

$((300*343) - 300v)/(290*343-145v) = 1$

Solving for v, we have,

$v = 22.12m/s$

Therefore the speed of the train before and after slowing down is 22.12m/s

Final answer:

The speed of the train can be determined using the Doppler effect formula.

Explanation:

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))

Where:

f' is the observed frequency
f is the source frequency
v is the speed of sound
v_o is the speed of the observer (here it is the train)
v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

Learn more about Doppler effect here:

brainly.com/question/33454469

#SPJ3

A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

Answers

Answer:

19.3 m/s

Explanation:

Take down to be positive. Given:

Δy = 19 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)

v = 19.3 m/s

A positively charged rod is held near a neutral conducting sphere as illustrated below. A positively charged particle is moved from point A to point B. Thee electrostatic work done on the positively charged particle during the motio

Answers

The movement of a positively charged particle from point A to point B. the motion-induced electrostatic work done on the positively charged particle.

Whether positively or negatively charged, an object that is neutral will interact with it in a pleasing way. Both positively charged and neutral items attract one another, as do negatively charged and neutral objects. These electrons gather on the further surface of sphere B, depleting the electron supply in sphere A. Therefore, sphere A (which is closer to the rod) obtains a positive charge and sphere B acquires a negative charge when the two spheres separate in the presence of the rod. The change in the particle's electrostatic potential energy in the external field equals the work done by the external force. When a charge is pushed from point A to point B, its potential energy changes, representing the labor of an outside force.

to know more about electrostatic work done please visit.

brainly.com/question/14947575

#SPJ4

If a pressure gauge measure an increase in 3×10^(5)Pa on an area of 0.7 m^2, calculate the increase in the force applied to the area?

Answers

Answer:210000N

Explanation:

Pressure=3x10^5pa

area=0.7m^2

Force = pressure x area

Force=3x10^5x0.7

Force=210000N

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min

Answers

Answer:

T = 188.5 s, correct is C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

L_f = I₀ w + m r v

L₀ = L_f

m r v₀ = I₀ w + m r v

angular and linear velocity are related

v = r w

w = v / r

m r v₀ = I₀ v / r + m r v

m r v₀ = (I₀ / r + mr) v