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nadya68 [22]
2 years ago
13

A heat engine extracts 42. 53 kj from the hot reservior and exhausts 17. 69 kj into the cold reservior. what is the work done?

Physics
1 answer:
ludmilkaskok [199]2 years ago
6 0

The answer is  24.84kJ.

We apply the expression for the work done by the heat engine is,

W=E_{i n}-E_{o u t}. Putting all given values in the equation we get the final answer.

What is heat engine?

  • A heat engine is a machine that uses heat to generate power. It draws heat from a reservoir, uses that heat to produce work, such as move a piston or lift weights, and then releases that heat energy into the sink.
  • We are given:The heat input is $E_{i n}=42.53 \mathrm{~kJ}$. The heat output is $E_{o u t}=17.69 \mathrm{~kJ}$.
  • The expression for the work done by the heat engine is,W=E_{i n}-E_{o u t}
  • Substituting the given values in the above expression, we will getW =42.53 \mathrm{~kJ}-17.69 \mathrm{~kJ}=24.84kJ.
  • Thus, the work done by the heat engine is 24.84kJ.

To learn more about heat engine visit: brainly.com/question/15735984

#SPJ4

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Heat is extracted from a certain quantity of steam at
vodomira [7]

Answer:v=2452.91 m/s

Explanation:

Given

initially steam is at 100^{\circ}C and converted to 0^{\circ} C ice

Let m be the mass of steam

latent heat of fusion and vaporization for water is

L_f=3.33\times 10^5 J/kg

L_v=2.26\times 10^6 J/kg

Heat required to convert steam in to water at 100^{\circ}C

Q_1=m\times L_v=m\cdot 2.26\times 10^6 J

Heat required to lower water temperature to 0^{\circ}C

Q_2=m\times c\times \Delta T

Q_2=m\times 4.184\times (100)

Q_2=4.184m\times 10^5 J

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Q_3=m\times L_f

Q_3=m\times 3.33\times 10^5=3.33m\times 10^5 J

Q=Q_1+Q_2+Q_3

Q=(2.26+0.4184+0.33)m\times 10^6 J

Q=3.0084m\times 10^6 J

So this energy is equal to kinetic energy of  bullet of mass m moving with velocity v

Q=\frac{1}{2}mv^2

3.0084m\times 10^6=\frac{1}{2}mv^2

v^2=3.0084\times 2\times 10^6

v=2.452\times 10^3 m/s

v=2452.91 m/s  

5 0
3 years ago
What can be correctly concluded from this experiment?
solong [7]
1


Step by step explanation: 1

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3 years ago
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QC Suppose you hear a clap of thunder 16.2s after seeing the associated lightning strike. The speed of light in air is 3.00× 10⁸
MAVERICK [17]

If you hear a clap of thunder in a time of 16.2s after seeing the associated lightning strike, you are: 5508 m far from the lightning strike

To solve this problem we must consider that the speed of light is greater than the speed of sound, therefore to calculate the distance we must use the speed of sound (340 m/s).

The formula and procedure we will use to solve this exercise is:

x = v * t

Where:

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Information about the problem:

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Applying the distance formula we have that:

x = v * t

x= 340 m/s * 16.2 s

x = 5508 m

<h3>What is velocity?</h3>

It is a physical quantity that indicates the displacement of a mobile per unit of time, it is expressed in units of distance per time, for example (miles/h, km/h).

Learn more about velocity at: brainly.com/question/80295?source=archive

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7 0
1 year ago
Need help ASAP!!!! here are the options: Amplitude, Compression, Rarefaction, and Wavelength
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What is the amplitude of a wave related to
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Answer:

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Explanation:

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