Entropy of a sealed room with an open-door refrigerator

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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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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.

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