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Arada [10]
3 years ago
9

How would a system which exchanges energy but does not exchange matter between the system and the surroundings be classified?

Physics
2 answers:
Elodia [21]3 years ago
4 0
If a system allows the exchange of energy but does not permit exchange of matter between the system and surroundings, this system is classified as a CLOSED system. The energy that flows in and out of the system boundaries is in the form of heat. 
Margaret [11]3 years ago
4 0

Explanation:

When a system exchanges only energy and there is no exchange of matter between the system and surrounding then this means the system is closed.

For example, when we placed hot water in a closed jar then after a certain hours the water inside the container becomes cold. This is because energy in the form of heat has been exchanges with the atmosphere or surroundings.

Hence, there is no exchange of matter between the system and surrounding.

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Sensory receptors pick up on senses from outside the body. To respond to the sense, what path does the information follow?
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Answer:

Follow that smell. SMELL: Originally thought to be present only in the nose, olfactory receptors exist in many locations in the body, including in the skin, heart, lungs, kidneys, muscles, and sperm. ...

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A taste for pathogens

Explanation:

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2 years ago
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Which is true of the greenhouse effect?
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A construction worker puts 20J of energy in to one strike of his hammer on the head of a nail. The energy transferred to driving
Slav-nsk [51]

Answer:

efficiancy=40 percent

Explanation:

efficiency=energy output/energy input×100

efficiancy=8J/20J×100

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Visible light includes all of the
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6 0
3 years ago
A gas expands against a constant external pressure of 2.00 atm until its volume has increased from 6.00 to 10.00 L. During this
mars1129 [50]

Answer:

ΔU = - 310.6 J (negative sign indicates decrease in internal energy)

W = 810.6 J

Explanation:

a.

Using first law of thermodynamics:

Q = ΔU + W

where,

Q = Heat Absorbed = 500 J

ΔU = Change in Internal Energy of Gas = ?

W = Work Done = PΔV =

P = Pressure = 2 atm = 202650 Pa

ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³

Therefore,

Q = ΔU + PΔV

500 J = ΔU + (202650 Pa)(0.004 m³)

ΔU = 500 J - 810.6 J

<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>

<u></u>

b.

The work done can be simply calculated as:

W = PΔV

W = (202650 Pa)(0.004 m³)

<u>W = 810.6 J</u>

7 0
3 years ago
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