Entropy of a sealed room with an open-door refrigerator

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In a sealed room with an open-door refrigerator, the entropy of the system increases due to the energy dynamics involved. While the refrigerator cools its interior, it simultaneously releases heat into the room, leading to a net increase in the room's entropy. The processes of compression and expansion within the refrigerator are adiabatic, but the system as a whole is not isolated; work is done to circulate the refrigerant, contributing to the overall energy increase. The discussion highlights that even though the heat exchange seems balanced, the energy input from the refrigerator's operation leads to an increase in entropy. Therefore, the correct answer is that the entropy increases, contrary to the initial assumption.
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Homework Statement
A working refrigerator with the door open is placed in a sealed room.

What change occurs to the entropy of the room?
Relevant Equations
Q= ΔU + W
dS = dQ / T
A working refrigerator with the door open is placed in a sealed room.
The entropy of the room
A. is zero.
B. decreases.
C. remains unchanged.
D. increases.

I chose C.
Here's my thought process:
In a working refrigerator, a compressor compresses a refrigerant (at a gas state) in the coil, and the refrigerant turns into liquid. The latent heat of condensation is rejected into the room, making room hotter. When the refrigerant travels into the refrigerator, an expansion valve expands the refrigerant and converts it into gas. The latent heat of fusion comes from the interior of the refrigerator, which causes the inside to become cooler. However, in this case, the door is open. This means that the net heat exchange is 0 as latent heat of condensation = latent heat of fusion.

In my knoweldge, the compression and expansion processes are adiabatic, so there is no heat transferred to or from the system.

But the mark scheme says the entropy increases. Can you explain which part of my solution is wrong and why?
 
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Is it plugged in?
 
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it says 'working'
 
techsingularity2042 said:
it says 'working'
Right, so think about the energy.
 
techsingularity2042 said:
it says 'working'
What @haruspex is saying is that you've gotten lost in the details. Take as your system the combination of room (assumed insulated) and refrigerator. Is work being done on that system? Is the internal energy of that system increasing?
 
techsingularity2042 said:
In my knowledge, the compression and expansion processes are adiabatic, so there is no heat transferred to or from the system.

But the mark scheme says the entropy increases. Can you explain which part of my solution is wrong and why?
What energizes the compression process, which must overcome the friction resistance of the tubes-condenser-evaporator to circulate through the system and to go through the expansion orifice or valve?
It seems to me that the "sealed room" is still being penetrated by some kind of external energy.

Fig.-6-6_revised-768x630.png
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

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