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A piston–cylinder device initially contains 2 L of air at 100 kPa and 25°C. Air is now compressed to a final state of 600 kPa an

bagirrra123 [75]

Answer:

a. The energy of the air at the initial and the final states is 0kJ and 0.171kJ respectively

b. 0.171kJ

c. 0.143

Explanation:

a.

Because there are same conditions of the state of air at the surroundings and at the Initial stage, the energy of air at the Initial stage is 0kJ.

Calculating energy at the final state;

We start by calculating the specific volume of air in the environment and at the final state.

U2 = At the final state, it is given by

RT2/P2

U1= At the Initial state, it is given by

RT1/P1

Where R = The gas constant of air is 0.287 kPa.m3/kg

T2 = 150 + 273 = 423K

T1 = 25 + 273 = 298K

P2 = 600KPa

P1 = 100KPa

U2 = 0.287 * 423/600

U2 = 0.202335m³/kg

U1 = 0.287 * 298/100

U1 = 0.85526m³/kg

Then we Calculate the mass of air using ideal gas relation

PV = mRT

m = P1V/RT1 where V = 2*10^-3kg

m = 100 * 2 * 10^-3/(0.287 * 298)

m = 0.00234kg

Then we calculate the entropy difference, ∆s. Which is given by

cp2 * ln(T2/T1) - R * ln(P2/P1)

Where cp2 = cycle constant pressure = 1.005

∆s = 1.005 * ln (423/298) - 0.287 * ln(600/100)

∆s = -0.1622kJ/kg

Energy at the final state =

m(E2 - E1 + Po(U2 - U1) -T0 * ∆s)

E2 and E1 are gotten from energy table as 302.88 and 212.64 respectively

Energy at the final state

= 0.00234(302.88 - 212.64 + 100(0.202335 - 0.85526) - 298 * -0.1622)

Energy at the final state = 0.171kJ

b.

Minimum Work = ∆Energy

Minimum Work = Energy at the final state - Energy at the initial state

Minimum Work = 0.171 - 0

Minimum Work done = 0.171kJ

c. The second-law efficiency of this process is calculated by ratio of meaningful and useful work

= 0.171/1.2

= 0.143

3 0

3 years ago

How does a fuse work?

Vilka [71]

Answer:

A. a material burns out when current is excessive

5 0

3 years ago

A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

Lorico [155]

Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

Required

Acceleration of the cart a

To get that, we will use the equation of motion

v = u+at

Substitute the given parameters

8.2 = 3.5+1.5a

1.5a = 8.2-3.5

1.5a = 4.7

a = 4.7/1.5

a = 3.13cm/s²

Hence the acceleration to the cart is 3.13cm/s²

3 0

3 years ago

A person runs 15 kilometers in two hours what is his/her average speed

bazaltina [42]

Answer:

$\displaystyle V=7.5\ km/h$

Explanation:

<u>Average Speed </u>

If an object travels a distance d in a time t regardless of the direction, the average speed is the quotient of the distance over the time:

$\displaystyle V=\frac{d}{t}$

It's known a person runs d=15 kilometers in t=2 hours, thus his/her average speed is:

$\displaystyle V=\frac{15\ km}{2\ h}$

Calculating:

$\boxed{\displaystyle V=7.5\ km/h}$

5 0

3 years ago

The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earthâ€™s surface.

aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = $\frac{1}{3}$ g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = $\frac{1}{3}$ m * g₁ = $\frac{1}{3}$ w₁

⇒Weight of probe on Mars = $\frac{1}{3}$ Weight of probe on earth

6 0

4 years ago