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miss Akunina [59]
3 years ago
13

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

the entropy change of all the components of this engine and determine if it is completely reversible. How much total work does it produce?
Physics
1 answer:
diamong [38]3 years ago
4 0

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

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sveticcg [70]

Answer:

  about 19.6° and 73.2°

Explanation:

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  y = -4.9(x/s·sec(α))² +x·tan(α)

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The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

_____

I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

5 0
2 years ago
At a hot air balloon race, a person on the ground shots a ball of confetti from a cannon to start the race. One of the balloons
Sati [7]

Answer:

a) The balloon and the ball will meet after 1.43 s.

b) The ball will reach the balloon at 15.7 m above the ground.

Explanation:

The height of the confetti ball is given by the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

The height of the ball is given by this equation:

y = y0 + v · t

Where v is the constant velocity.

When the ball and the ballon meet, both heights are equal. Let´s consider the ground as the origin of the frame of reference so that y0 = 0:

y balloon = y ball

y0 + v · t = y0 + v0 · t + 1/2 · g · t²                  (y0 = 0)

11 m/s · t = 18 m/s · t -1/2 · 9.8 m/s² · t²

0 = -4.9 m/s² · t² + 18 m/s · t - 11 m/s · t

0 = -4.9 m/s² · t²  + 7 m/s · t

0 = t( -4.9 m/s² · t  + 7 m/s)

t = 0 and

0 = -4.9 m/s² · t  + 7 m/s

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t = 1.43 s

They will meet after 1.43 s

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y = 11 m/s · 1.43 s = 15.7 m

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7 0
3 years ago
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<h2>Hello</h2>

The answer is:

35000Kg.\frac{m}{s}=35000N

<h2>Why?</h2>

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p=m*v

Where:

m=mass\\v=velocity

So, calculating we have:

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Remember,

1N=1Kg.\frac{m}{s}

Have a nice day!

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