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4 years ago

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

Engineering

1 answer:

Natali5045456 [20]4 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 differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistance

yuradex [85]

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

$\frac{Vout}{V2 - V1 } = \frac{R2}{R1}$

= Voltage gain = R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

4 0

3 years ago

Write the formula for calculating mechanical advantage for Levers, pulleys and wheels and axles.

NemiM [27]

Answer:

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7 0

2 years ago

Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon

balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

$torque =\frac{power}{2\pi N}$

$power=2\pi N\times T$

P= $2\times \pi \times2500 \times 4.5$

P=70,685W

P=70.685 KW

power= $\frac{work done}{time}$

work done = power * time

= 70.685*30=2120.55J

= 2.12 kJ

7 0

4 years ago

Why is concrete on its own not a good material to use

Inga [223]

Answer:

It has poor tensile strength despite having high compressive strength

Explanation:

Concrete exhibits high compressive strength when used. However, it has very low compressive strength. This is the reason why concrete is normally combined with steel to make a composite building material called reinforced concrete. The steel reinforces concrete hence increasing the tensile strength in RC buildings. The end composite is durable and fireproof. Generally, the main reason why concrete is not use on its own is due to its poor tensile strength.

7 0

3 years ago

A particular DSL modem operates at 768 kbits/sec. How many bytes can it receive in 1 minute? USB 3.0 can send data at 5 Gbits/se

igor_vitrenko [27]

Answer:

a. 1.6 Kbytes/min

b. 10.417 Mbytes/min

Explanation:

a. The DSL modem operates at 768 Kbits/sec.

But,

8 bits = 1 byte and 1 Kbit = 1 000 bits, so that:

= $\frac{768 000}{8}$

= 96 000 bytes

Therefore, the modem operates at 96 Kbytes/sec.

The byte to be received in 1 minute can be calculated thus;

Since 60 seconds = 1 minute, then:

= $\frac{96000}{60}$

= 1600

= 1.6 Kbytes/min

The modem receives 1.6 Kbytes/min

b. The USB sends 5 Gbits/sec.

But, 8 bits = 1 byte and 1 Gbit = 1000000000 bits so that:

= $\frac{5000000000}{8}$

= 625000000

= 0.625 Gbytes

The USB sends 0.625 Gbyte/sec.

Since 60 seconds = 1 minute, then:

= $\frac{625000000}{60}$

= 10416666.67

= 10.417 Mbytes/min

3 0

3 years ago