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faltersainse [42]

3 years ago

10

If you know that the change in entropy of a system where heat was added is 12 J/K, and that the temperature of the system is 250

K, what is the amount of heat added to the system? a)-5J b)-125J c)- 600 J d)-5000 J e)-8000 J

1 answer:

SVEN [57.7K]3 years ago

3 0

Solution:

Given:

Change in entropy of the system, ΔS = 12J/K

Temperature of the system, $T_{o}$ = 250K

Now, we know that the change in entropy of a system is given by the formula:

ΔS = $\frac{\Delta Q}{T_{o}}$

Amount of heat added, ΔQ = $\Delta S\times T_{o}$

ΔQ = 3000J

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A distillation column is initially designed to separate a mixture of toluene and xylene at around ambient temperature (say, 100°

weeeeeb [17]

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Purchase cost= 87056

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Explanation:

solution is attached below

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3 years ago

2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the

Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

{

private long[] a; // ref to array a

private int nElems; // number of data items

//-----------------------------------------------------------

public HighArray(int max) // constructor

{

a = new long[max]; // create the array

nElems = 0; // no items yet

}

//-----------------------------------------------------------

public setValue find(long searchKey)

{ // find specified value

int j;

for(j=0; j<nElems; j++) // for each element,

if(a[j] == searchKey) // found item?

break; // exit loop before end

if(j == nElems) // gone to end?

return false; // yes, can't find it

else

return true; // no, found it

} // end find()

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element,

System.out.print(a[j] + " "); // display it

System.out.println("");

}

//-----------------------------------------------------------

} // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

HighArray arr; // reference to array

arr = new HighArray(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

int searchKey = 35; // search for item

if( arr.find(searchKey) )

System.out.println("Found " + searchKey);

else

System.out.println("Can't find " + searchKey);

} // end main()

} // end class HighArrayApp

Explanation:

6 0

2 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a

AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2) + x(2) * h(fg2)

so

exit quality = $\frac{h1 - h(f2)}{h(fg2)}$

exit quality = $\frac{332.17- 9.04}{1382.73)}$

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0

3 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$ .

True Strain

A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$ .

The mathematical relation between stress to strain on the plastic region of deformation is

$\sigma _T =K\epsilon^n_T$

Where K is a constant

n is known as the strain hardening exponent

This constant K can be obtained as follows

$K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

$K = \frac{345}{(0.02)^{0.22}}$

$= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

$\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

$\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

$=0.0443$

From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

$l_i = l_o e^{\epsilon_T}$

Where

$l_i$ is the instantaneous length

$l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

$l_i = 470 * e^{0.0443}$

$=491.28mm$

We can also obtain the elongated length mathematically as follows

$Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

$Elongated \ Length = 491.28 - 470$

$=21.29mm$

4 0

3 years ago