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Zinaida [17]
3 years ago
8

Assuming standard pressure, what physical constants are needed to calculate the change in enthalpy for raising a sample of water

from 250 K to 300 K. I) the heat capacity of ice II) the enthalpy of fusion of water III) the heat capacity of water IV) the enthalpy of vaporization of water V) the heat capacity of steam
Physics
1 answer:
Alik [6]3 years ago
8 0

Answer:

I) Heat capacity of ice (Sensible, ice), II) Enthalpy of fusion of water (Latent, fusion) and III) Heat capacity of water (Sensible, water).

Explanation:

At standard pressure, water has a fusion point and boiling point of 273.15 K and 373.15 K, respectively. Then, the total change in enthalpy is the sum of two sensible indicators and one latent indicator. That is:

\Delta H_{total} = \Delta H_{sensible, ice} + \Delta H_{latent, fusion} + \Delta H_{sensible, water}

The needed physical constants are: Heat capacity of ice (Sensible, ice), Enthalpy of fusion of water (Latent, fusion) and Heat capacity of water (Sensible, water).

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Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of
Andre45 [30]
Well first of all, I think the students may have been correct. 
If they didn't use distilled water, and if it wasn't exactly at 
standard temperature, then the mass of  25.0 mL  could
very well be  25.4 grams.  We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was  (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

                            +0.4 / 25.0  =  +0.016 .

To change a decimal to a percent, move the
decimal point two places that way  ===> .

                           + 0.016  =  +1.6 % .

     
Their measurement was 1.6% too high.

Let's not call it an 'error'.  Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value.  OK ?
4 0
3 years ago
Read 2 more answers
Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/sthrougha horizontal tube with a 30mmdiameter. what
marusya05 [52]

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>
  • We have the Poiseuille's formula,

                     Q=\frac{\pi r^4P}{8\beta l}

  • where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.
  • By substituting values and rearranging we will get the pressure drop as,

                  P=3.284Pascals

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.

Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

#SPJ4

3 0
1 year ago
Please help I need this done within 30 mins
earnstyle [38]

It may be thinner and more dense? I’m not too experienced in the study of Earth’s crust. However, I know enough to remember that the earths crust is thin.

7 0
2 years ago
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