Ask question

Ask question

All categories

3 years ago

8

A low-temperature physicist wishes to publish his experimental result that the heat capacity of a nonmagnetic dielectric materia

l between 0.05K and 0.5K varies as
A√ T + BT^3

As editor of the journaL should you accept the paper for publication?

1 answer:

tatiyna3 years ago

6 0

Answer:

Yes, this because the specific heat does not violate the third law of thermodynamics.

Explanation:

The third law of thermodynamic is usually used for a closed system to relate the thermodynamic properties of the system at equilibrium conditions. For this law, a body that exists at a temperature of 0 K will cease to move or stop moving. It can be inferred that heat capacity at (T = 0 K) is equivalent to 0. Therefore, the equation is valid for the given temperature range.

You might be interested in

An isentropic steam turbine processes 2 kg/s of steam at 3 MPa, which is exhausted at50 kPa and 100C. Five percent of this flow

borishaifa [10]

Answer:

2285kw

Explanation:

since it is an isentropic process, we can conclude that it is a reversible adiabatic process. Hence the energy must be conserve i.e the total inflow of energy must be equal to the total outflow of energy.

Mathematically,

$\\ E_{inflow} = E_{outflow}$

Note: from the question we have only one source of inflow and two source of outflow (the exhaust at a pressure of 50kpa and the feedwater at a pressure of 5ookpa). Also the power produce is another source of outgoing energy $\\ E_{inflow} = m_{1} h_{1}$ .

$\\$

$E_{outflow} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$

$\\$

Where $m_{1} h_{1}$ are the mass flow rate and the enthalpies at the inlet at a pressure of 3Mpa $\\$ ,

$m_{2} h_{2}$ are the mass flow rate and the enthalpies at the outlet 2 where we have a pressure of 500kpa respectively. $\\$ ,

and $m_{3} h_{3}$ are the mass flow rate and the enthalpies at the outlet 3 where we have a pressure of 50kpa respectively. $\\$ ,

We can now express write out the required equation by substituting the new expression for the energies $\\$

$m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$

from the above equation, the unknown are the enthalpy values and the mass flow rate. $\\$

first let us determine the enthalpy values at the inlet and the out let using the Superheated water table. $\\$

It is more convenient to start from outlet 3 were we have a temperature $100^{0}C$ and pressure value of (50kpa or 0.05Mpa ). using double interpolation method on the superheated water table to determine the enthalpy value with careful calculation we have $\\$

$h_{3}$ = 2682.4 KJ/KG , at this point also from the table the entropy value , $s_{3}$ value is 7.6953 KJ/Kg.K. $\\$

Next we determine the enthalphy value at outlet 2. But in this case, we don't have a temperature value, hence we use the entrophy value since the entropy is constant at all inlet and outlet. $\\$

So, from the superheated water table again, at a pressure of 500kpa (0.5Mpa) and entropy value of 7.6953 KJ/Kg.K with careful interpolation we arrive at a enthalpy value of 3206.5KJ/Kg. $\\$

Finally for inlet one at a pressure of 3Mpa, interpolting with an entropy value of 7.6953KJ/Kg.K we arrive at enthalpy value of 3851.2KJ/Kg. $\\$

Now we determine the mass flow rate at each inlet and outlet. since mass must also be balance, i.e $m_{1} = m_{2} + m_{3}$ $\\$

From the question the, the mass flow rate at the inlet $m_{1}}$ is 2Kg/s $\\$

Since 5% flow is delivered into the feedwater heating, $\\$

$m_{2} = 0.05m_{1} = 0.05 *2kg/s = 0.1kg/s$ $\\$

Also for the outlet 3 the remaining 95% will flow out. Hence

$m_{3} = 0.95m_{1} = 0.95 *2kg/s = 1.9kg/s$ $\\$

Now, from $m_{1} h_{1} = m_{2} h_{2} + m_{3} h_{3} + W_{out}$ $\\$ we substitute values

$W_{out} = m_{1} h_{1}-m_{2} h_{2}-m_{3} h_{3}$

$W_{out} = (2kg/s)(3851.2KJ/Kg) - (0.1kg/s)(3206.5kJ/kg)- (1.9)(2682.4kJ/kg)$

$\\$

$W_{out} = 2285.19 kW$ .

Hence the power produced is 2285kW

7 0

3 years ago

A soil weighs 2,520 lbs/CY in its in situ condition, 1,970 lb/CY in its loose condition after excavation, and 3,025 lbs/CY in an

azamat

Answer:

load factor = 0.782

Shrink Factor = 0.833

no of truck is 62500

Explanation:

given data

soil weighs in situ condition = 2,520 lbs/CY

soil weighs in loose condition = 1,970 lb/CY

soil weighs in embanked state = 3,025 lbs/CY

average volume = 16 LCY

soil from a borrow pit = 1 million CCY

solution

first we get here Load Factor that is express as

load factor = $\frac{1,970}{2550}$

load factor = 0.782

and Shrink Factor will be as

Shrink Factor = $\frac{2520}{3025}$

Shrink Factor = 0.833

and

no of truck will be

no of truck = $\frac{1000000}{16}$

no of truck is 62500

6 0

4 years ago

This method will sell the seat in row i and column j unless it is already sold. A ticket is sold if the price of that seat in th

blsea [12.9K]

Answer:

The solution code is written in Java.

public class Movie {
private double [][] seats = new double[5][5];
private double totalSales;
public Movie(){
for(int i= 0; i < this.seats.length; i++){
for(int j = 0; j < this.seats[i].length; j++){
this.seats[i][j] = 12;
}
}
this.totalSales = 0;
}
public boolean bookSeat(int i, int j)
{
if(this.seats[i][j] != 0){
this.totalSales += this.seats[i][j];
this.seats[i][j] = 0;
return true;
}else{
return false;
}
}
}

Explanation:

The method, bookSeat(), as required by the question is presented from Line 16 - 26 as part of the public method in a class <em>Movie</em>. This method take row,<em> i</em>, and column,<em> j</em>, as input.

By presuming the seats is an two-dimensional array with all its elements are initialized 12 (Line 7 - 10). This means we presume the movie ticket price for all the seats are $12, for simplicity.

When the<em> bookSeat() </em>method is invoked, it will check if the current price of seats at row-i and column-i is 0. If not, the current price, will be added to the <em>totalSales </em>(Line 19)<em> </em>and then set the price to 0 (Line 20) and return <em>true</em> since the ticket is successfully sold (Line 21). If it has already been sold, return <em>false</em> (Line 23).

8 0

4 years ago

1 british gallon = .....litres.<br>a:4.546<br>b:3.785<br>c.5.456<br>d.7.385

solmaris [256]

Answer:

c

Explanation:

I live in britian

8 0

3 years ago

Describe two components of the MEDC mechanized agricultural system that may lead to an increase in water stress

Zina [86]

Mechanized agriculture refers to agricultural machinery capable of facilitating and enhancing agriculture. Mechanized agriculture does not require moving the soils and therefore is hard for water to enter into it.

<h3>What is mechanized agriculture?</h3>

Mechanized agriculture refers to the use of different technologies (e.g., precision agriculture GPs) in order to increase farmworker productivity.

Mechanized agriculture is a common practice in developed countries (MEDCs), whereas this practice is relatively uncommon in developing countries (LEDCs).

The techniques of mechanized agriculture do not require moving the soils and therefore is hard for water to enter into it.

Learn more about mechanized agriculture here:

brainly.com/question/1543791

8 0

3 years ago