Is Environmental Nature or Nurture?

What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

or

$\frac{Q_2}{3.78\times 10^3}=0.72$

or

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

or

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

7 0

3 years ago

The ____ force is the sum of all the forces that act upon each other

Pavel [41]

the resultant force

5 0

3 years ago

Read 2 more answers

in a human cannonball a person is shot from a cannon with a barrel that is 3.05 m long. 16700 J of work are done to accelerate t

ankoles [38]

The work done to accelerate the acrobat is given by
$W=Fd$
where F is the force applied and d the distance of application of the force.
If the barrel is 3.05 m long, then d=3.05 m. Therefore we can find the force:
$F= \frac{W}{d} = \frac{17600 J}{3.05 m} =5475 N$

6 0

3 years ago

How do you calculate relative velocity?

nekit [7.7K]

Answer:

<h3>Vab=Va-Vb</h3><h3>Vab=Vb-Va</h3><h3>Explanation:</h3>

Vab is velocity of a relative to b

Vba is velocity of b relative to a

4 0

3 years ago

An object is place 30.5 cm from a convex lens that has a focal length of 8.0cm.where is the image ?

77julia77 [94]

Answer:

Image is between F and 2F. And the exact location is:

q = 10.84 cm

Explanation:

The convex lens form different kinds of images with different sizes and locations, for the different positions of an object. In this case, the object is placed 30.5 cm from the convex lens with a focal length of 8 cm. This means that it is the case of the object placed beyond 2F. In this case, the image formed has the following characteristics:

1. Image is real.

2. Image is inverted

3. Image is diminished

<u>4. Image is located between F and 2F</u>

For exact location we use the thin lens formula:

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where,

f = focal length = 8 cm

p = object distance = 30.5 cm

q = image distance = ?

Therefore,

$\frac{1}{8\ cm} = \frac{1}{30.5\ cm} + \frac{1}{q} \\\frac{1}{8\ cm} - \frac{1}{30.5\ cm} = \frac{1}{q}$

7 0

2 years ago