All categories

3 years ago

Explain what entropy is in relation to the second law of thermodynamics?

Engineering

1 answer:

Marat540 [252]3 years ago

5 0

Explanation:

Second law tell us that ,the entropy of reversible process is remain constant while the entropy of ir-reversible process increases.

Entropy of universe is the summation of entropy of system and entropy of surrounding.So we can say that

$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surrounding}$

Entropy can be define as

$ds=\int \dfrac{dQ}{T}$

Entropy is point function.This is property of system.Ir-reversibility cause the increase of entropy.

$\oint \dfrac{dQ}{T} =0 ,this\ is\ reversible\ process$

$\oint \dfrac{dQ}{T} >0 ,this\ is\ impossible\ process$

$\oint \dfrac{dQ}{T}$

You might be interested in

Conclusion. What process is responsible for the bubbling action of the organism? What is your evidence?

noname [10]

Answer:

Explanation:

Hands-on Activity Bubbling Plants Experiment to Quantify Photosynthesis ... After running the experiment, students pool their data to get a large sample ... Explain that photosynthesis is a process that plants use to convert light ... Describe a simple experiment that provides indirect evidence that photosynthesis is occurring.

Through photosynthesis, certain organisms convert solar energy (sunlight) into ... of our planet continuously and is transferred from one organism to another. Therefore, directly or indirectly, the process of photosynthesis provides most of the energy ... Biology in Action ... Chlorophyll is responsible for the green color of plants.Photosynthetic organisms capture energy from the sun and matter from the air to ... oxygen produced during photosynthesis makes leaf bits float like bubbles in water. ... their ability to carry out photosynthesis, the biochemical process of capturing ... this air is forced out and replaced with solution, causing the leaves to sink.

6 0

3 years ago

5. What number filter lens is recommended when welding with SMAW and using 1/8" (3.2 mm)

Inessa05 [86]

Answer:

What filter lens is recommended for SMAW?

Table of Filter Lenses for Protection From Radiant Energy

Type of Operation Electric Size 1/32 in. Arc Current (Amp)

Shield metal arc welding (SMAW) 3 to 5 60 to 160

5 to 8 160 to 250

More than 8 250 to 500

Explanation:

7 0

3 years ago

For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of allo

Zina [86]

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= $\frac{R}{R +S } * 100$

= $\frac{53-45}{58-45} * 100$ = 61.5%

To determine the wt.% of the alloy that is liquid

= $\frac{S}{S+R} * 100\\$

= $\frac{58-53}{58-45} *100$ = 38.5%

attached below is a free hand sketch as well

7 0

3 years ago

Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr

maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

$P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}$

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

$\eta_t$ = The efficiency of the pump = 78%

$\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}$

$Q_{max}$ = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump $Q_{max}$ ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

6 0

3 years ago

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures befo

SVEN [57.7K]

This question is incomplete, the complete question is;

An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.

Answer:

a) the maximum pressure in the cycle is 30.01 Mpa

b) the heat transfer to air is 0.7058 KJ

c) mass of Air is 0.002957 kg

Explanation:

Given the data in the question;

We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)

so from the "ideal gas properties of air table"

Pr1 = 238

Pr4 = 2.379

we know that Pressure P1 is only maximum at the beginning of the expansion process,

now we express the relative pressure and pressure relation for the process 4-1

P1 = (Pr2/Pr4)P4

so we substitute

P1 = (238/2.379)300 kPa

P1 = 30012.6 kPa = 30.01 Mpa

Therefore the maximum pressure in the cycle is 30.01 Mpa

the Thermal heat efficiency of the Carnot cycle is expressed as;

ηth = 1 - (TL/TH)

we substitute

ηth = 1 - (350K/1200K)

ηth = 1 - 0.2916

ηth = 0.7084

now we find the heat transferred

Qin = W_net.out / ηth

given that the net work output per cycle is 0.5 kJ

we substitute

Qin = 0.5 / 0.7084

Qin = 0.7058 KJ

Therefore, the heat transfer to air is 0.7058 KJ

first lets express the change in entropy for process 3 - 4

S4 - S3 = (S°4 - S°3) - R.In(P4/P3)

S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa

= -0.1989 Kj/Kg.K = S1 - S2

so that; S2 - S1 = 0.1989 Kj/Kg.K

Next we find the net work output per unit mass for the Carnot cycle

W"_netout = (S2 - S1)(TH - TL)

we substitute

W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K

= 169.065 kJ/kg

Finally we find the mass

mass m = W_ net.out / W"_netout

we substitute

m = 0.5 / 169.065

m = 0.002957 kg

Therefore, mass of Air is 0.002957 kg

5 0

3 years ago