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3 years ago

Explain the effect of heat transfer, reversibilities and irreversibilities on changes in entropy

Engineering

1 answer:

Natali5045456 [20]3 years ago

5 0

Answer is given below

Explanation:

The second law of thermodynamics states that in the reversible process, the entropy of the universe is constant, while in the unchanging process, such as the transfer of heat from a hot object to a cold object, the reflection of the universe increases.

ΔS = ΔQ ÷ T

The irreversible process increases the entropy of the universe. Since entropy is a state function, the change in entropy of the system is the same, the process is reversible or irreversible. The second law of thermodynamics can be used to determine whether a process is reversible.

for an irreversible process, ΔS is not same as reversible process, because more than 1 reversible process is required for 2 ΔS to be equal.

if the initial and final states are the same, ΔS for an irreversible process can be calculated as if it were a reversible process.

If the initial and final condition is same then ΔS can find out for irreversible process even if it were reversible process

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A satellite would have a mass of 270 kg on the surface of Mars. Determine the weight of the satellite in pounds if it is in orbi

koban [17]

Answer:

26 lbf

Explanation:

The mass of the satellite is the same regardless of where it is.

The weight however, depends on the acceleration of gravity.

The universal gravitation equation:

g = G * M / d^2

Where

G: universal gravitation constant (6.67*10^-11 m^3/(kg*s))

M: mass of the body causing the gravitational field (mass of Earth = 6*10^24 kg)

d: distance to that body

15000 miles = 24140 km

The distance is to the center of Earth.

Earth radius = 6371 km

Then:

d = 24140 + 6371 = 30511 km

g = 6.67*10^-11 * 6*10^24 / 30511000^2 = 0.43 m/s^2

Then we calculate the weight:

w = m * a

w = 270 * 0.43 = 116 N

116 N is 26 lbf

8 0

3 years ago

How can any student outside apply for studying engineering at Cambridge University

telo118 [61]

Admission to the Engineering course at Cambridge is highly competitive, both in terms of the numbers and quality of applicants. In considering applicants, Colleges look for evidence both of academic ability and of motivation towards Engineering. There are no absolute standards required of A Level achievement, but it should be noted that the average entrant to the Department has three A* grades. You need to get top marks in Maths and Physics.All Colleges strongly prefer applicants for Engineering to be taking a third subject that is relevant to Engineering.
Hope that helps and good luck if you are applying. Can you please mark this as brainliest and press the thank you button and if you have any further questions please let me know!!

3 0

3 years ago

One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter

lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150× $10^{-3}$ $m^{3}$

m = 1 Kg

$P_{1}$ = 2 MPa

$T_{2}$ = 40°C

The waters specific volume is calculated:

$v_{1}$ = V/m

Here, the waters specific volume at initial condition is $v_{1}$ , the containers volume is V, waters mass is m.

$v_{1}$ = 150× $10^{-3}$ $m^{3}$ /1

$v_{1}$ = 0.15 $m^{3}$ / Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 $m^{3}$ / Kg and 0.13 $m^{3}$ / Kg.

$T_{1}$ = 350+(400-350) $\frac{0.15-0.13}{0.1522-0.1386}$

$T_{1}$ = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

$v_{2}$ = $v_{1}$ = 0.15 $m^{3}$ / Kg

In saturated water labels for $T_{2}$ = 40°C.

$v_{f}$ = 0.001008 $m^{3}$ / Kg

$v_{g}$ = 19.515 $m^{3}$ / Kg

The final state is two phase region $v_{f}$ < $v_{2}$ < $v_{g}$ .

In saturated water labels for $T_{2}$ = 40°C.

$P_{2}$ = $P_{Sat}$ = 7.3851 kPa

$P_{2}$ = 7.3851 kPa

7 0

3 years ago

compressors, the gas is often cooled while being compressed to reduce the power consumed by the compressor. explain how cooling

ASHA 777 [7]

The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.

When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.

Learn more about thermodynamics here-

brainly.com/question/1368306

#SPJ4

8 0

1 year ago

To reduce the negative impacts of stress on your mental

soldi70 [24.7K]

Answer:

A&C

Explanation:

breathing deeply is relaxing

talking with a friend can helping