Help Understanding Thermo Principles

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The discussion revolves around understanding thermodynamic principles, specifically the energy analysis of a closed system using the equation deltaU = Q - W. The user is confused about the role of work done by a fan in a scenario where heat transfer is disregarded, leading to the conclusion that the room's temperature will not decrease. Despite calculating internal energy changes using deltaU = mass * Cv * (T2 - T1), the user questions why the work done by the fan is considered internal energy rather than external work. The conversation highlights the importance of grasping energy balance concepts in thermodynamics, emphasizing that clarity in definitions is crucial for problem-solving. Ultimately, the user seeks a deeper understanding of how work and internal energy interact in this context.
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Okay, I need some help understanding some of the Thermo principles and I found this forum. I hope someone can help.

I'm working on Energy Analysis of a closed system. Conservation of energy principle and all that. deltaU = Q-W. Every time I think I know what should be considered work and what should be considered internal energy (U), etc. I am proven wrong.

In one example we have a room where we are disregarding any heat transfer through walls and doors. We are given a constant room volume, an initial Pressure and Temperature, the Wattage of a running fan and an amount of Time that the fan is running for. We are asked if the room cools down,which of course it will not because no heat is being removed and the work from the fan will add heat to the room.

Q=0 and we are left with deltaU = -W
delta U = Wattage * Time = kJ
that is one thing that confuses me because I would have thought the work by the fan would = Work or negative work whatever the case might be.

The problem boils down to
delta U = mass * Cv * (T2-T1) and we solve for T2
(mass and Cv were calculated for air from data given)

What happened to work? I'm confused. :frown:
 
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Just, balance, energy, period.

\dot {E_{in}} - \dot{E_{out}} = \frac {dE}{dt} |_{cv}

VIdt - work-energy in

Q_{in} - heat flow in (assumed)

m2u2 - m1u2, change in energy of the control volume.

Don't plug and chug. Learn what you are doing.
 
Clearly, I'm not plugging and chugging. I've worked through the entire problem and I'm asking a logic question. I can see why you are using the energy balance formula, however, it does not address my question.

My question is - why is the work by the fan being considered internal energy and not work? In this problem Q=0 so I'm not sure what your reference to Q is.
 
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why is the work by the fan being considered internal energy and not work

Did you read what I wrote for you?

me said:
VIdt - work-energy in

Does that say internal energy?

Ok, Q=0. In any event, it will tell you Q=0 when you solve the problem.
 
Thermo

I guess I'm going to get more attitude than help here.
 
You are the one who made the comment about blanket statements when tried to help you. In any event, let's start over.
 
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