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2 years ago

Is an isothermal process necessarily internally reversible? Explain your answer with an example

Engineering

1 answer:

torisob [31]2 years ago

5 0

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

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Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of subst

Kaylis [27]

Answer:

The specific heat capacity of substance A is 1.16 J/g

Explanation:

The substances A and B come to a thermal equilibrium, therefore, the heat given by the hotter substance B is absorbed by the colder substance A.

The equation becomes:

Heat release by Substance B = Heat Gained by Substance A

The heat can be calculated by the formula:

Heat = mCΔT

where,

m = mass of substance

C = specific heat capacity of substance

ΔT = difference in temperature of substance

Therefore, the equation becomes:

(mCΔT) of A = (mCΔT) of B

FOR SUBSTANCE A:

m = 6.01 g

ΔT = Final Temperature - Initial Temperature

ΔT = 46.1°C - 20°C = 26.1°C

C = ?

FOR SUBSTANCE B:

m = 25.6 g

ΔT = Initial Temperature - Final Temperature

ΔT = 52.2°C - 46.1°C = 6.1°C

C = 1.17 J/g

Therefore, eqn becomes:

(6.01 g)(C)(26.1°C) = (25.6 g)(1.17 J/g)(6.1°C)

C = (182.7072 J °C)/(156.861 g °C)

C = 1.16 J/g

5 0

3 years ago

is released from under a vertical gate into a 2-mwide lined rectangular channel. The gate opening is 50 cm, and the flow rate in

SVEN [57.7K]

Answer:

hello your question is incomplete attached below is the complete question

answer: There is a hydraulic jump

Explanation:

First we have to calculate the depth of flow downstream of the gate

y1 = $C_{c} y_{g}$ ----------- ( 1 )

Cc ( concentration coefficient ) = 0.61 ( assumed )

Yg ( depth of gate opening ) = 0.5

hence equation 1 becomes

y1 = 0.61 * 0.5 = 0.305 m

calculate the flow per unit width q

q = Q / b ----------- ( 2 )

Q = 10 m^3 /s

b = 2 m

hence equation 2 becomes

q = 10 / 2 = 5 m^2/s

next calculate the depth before hydraulic jump y2 by using the hydraulic equation

answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel

attached below is the remaining part of the solution

4 0

3 years ago

1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-

Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * $\frac{79.25}{21.2}$

= 0.0214 * 3.738

= 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0

3 years ago

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

5 0

3 years ago

Doubling the diameter of a solid, cylindrical wire doubles its strength in tension.

julsineya [31]

Answer:

True ❤️

-Solid by solid can make Cylindrical wire doubles Strengths in tension

4 0

3 years ago