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2 years ago

A system of releases 125kJ of heat while 104kJ of work is done in the system. Calcilate the change om imternal energy (in kJ)

Physics

1 answer:

artcher [175]2 years ago

4 0

Answer:

DU = 21 KJ

Explanation:

Given the following data;

Quantity of heat = 125 KJ

Work = 104 KJ

To find the change in internal energy;

Mathematically, the change in internal energy of a system is given by the formula;

DU = Q - W

Where;

DU is the change in internal energy.

Q is the quantity of energy.

W is the work done.

Substituting into the formula, we have;

DU = 125 - 104

DU = 21 KJ

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What animal did wolfgang kohler use to study insight learning

Katena32 [7]

Answer:

He used a chimp to study

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3 years ago

If hydrogen and helium are the most abundant elements in the unoverse , then why are living organisms composed primarily of carb

Mumz [18]

Answer:

Most of materia isnt life.

Explanation:

The living organisms (life) aren't the most abundant thing in universe.

Hydrogen and helium are present in everywhere, but life isn't.

There is no reason to think because we have a lot of a thing, the life must be made for this thing.

The organic life just can exists because some mysterious properties about carbon, that is the basic foundation of life, carbon is a special element, why? We don't know, actually, it's a huge problem for science discover why the carbon can makes life be possible and other elements can't. But we know is this element that makes life possible.

So, note there isn't relation about the quantity of a material in Universe and the life constituition. In addition, look around, organic materials are very rare in Universe, Earth is one in lots of places and in most of this places there isn't sign of life.

Even in Earth the life looks abundant, in Universe it isn't, the same way in Universe the Hydrogen and Helium are abudant, in Earth isn't soo.

3 0

3 years ago

An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,

Sedbober [7]

Answer:

$a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa$

Explanation:

a. Internal energy and the relative specific volume at $s_1$ are determined from A-17: $u_1=214.07kJ/kg, \ \alpha_r_1=621.2$ .

The relative specific volume at $s_2$ is calculated from the compression ratio:

$\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825$

#from this, the temperature and enthalpy at state 2, $s_2$ can be determined using interpolations $T_2=862K$ and $h_2=890.9kJ/kg$ . The specific volume at $s_1$ can then be determined as:

$\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg$

Specific volume, $s_2$ :

$\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg$

The pressures at $s_2 \ and\ s_3$ is:

$P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa$

.The thermal efficiency=> maximum temperature at $s_3$ can be obtained from the expansion work at constant pressure during $s_2-s_3$

$\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K$

b.Relative SV and enthalpy at $s_3$ are obtained for the given temperature with interpolation with data from A-17 : $a_r_3=4.553 \ and\ h_3=1909.62kJ/kg$

Relative SV at $s_4$ is

$a_r_4=\frac{r}{r_c}\alpha _r_3$

= $=\frac{16}{2}\times4.533\\=36.424$

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

$n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563$

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

$MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa$

Hence, the mean effective relative pressure is 674.95kPa

3 0

3 years ago

Type the correct answer in the box. use numerals instead of words. anne has a sample of a substance. its volume is 20 cm3, and i

CaHeK987 [17]

The density of sample is 5 g/cm3

Given:

volume of sample = 20 cm3

mass of sample = 100 grams

To Find:

density of sample

Solution: Density is the measure of how much “stuff” is in a given amount of space. For example, a block of the heavier element lead (Pb) will be denser than the softer, lighter element gold (Au). A block of Styrofoam is less dense than a brick. It is defined as mass per unit volume

density = mass/volume

d = 100/20

d = 5 g/cm3

So, density of sample is 5 g/cm3

Learn more about Density here:

brainly.com/question/1354972

#SPJ4

6 0

1 year ago

Find the mass of an object if a 40 N of force causes the object to accelerate at 5.5 m/s/s

Murrr4er [49]

Answer:

<h3>The answer is 8 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{40}{5} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago