Thermodynamics - example with answer

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Homework Help Overview

The discussion revolves around a thermodynamics problem related to the first law of thermodynamics, specifically focusing on the concepts of work done by a system and changes in internal energy. Participants are trying to clarify the implications of positive and negative work in the context of an expanding aluminium cube.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the interpretation of work as negative when done by the system and discussing the relationship between heat, internal energy, and work. There are inquiries about the use of temperature scales in the context of energy changes.

Discussion Status

The discussion is active, with participants exploring different interpretations of work in thermodynamic processes. Some guidance has been provided regarding the signs of work and the relationship between temperature differences, but no consensus has been reached on the specific examples being compared.

Contextual Notes

Participants are grappling with the sign conventions in thermodynamics and the implications of temperature measurements in Celsius versus Kelvin. There are references to external examples that may influence their understanding.

manal950
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Hi
Here is a question with answer , in fact I don't understand the answer fully ..
please can explain to me the answer .. pleas anyone clear out the answer

and here in answer I saw that work is negative is that because work done by the system ?

The question : -

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Answer

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It is based on the 1st law of TD which relates that the energy of the system is conserved.
If Q is the (+) heat that it receives then this energy can be used to raise the internal energy of the system ΔU (+) and/or to do work W (+) thus

Q = ΔU + W

so that

ΔU = Q - W

that is the difference in energy between the received heat and the work done by the system is used to raise its internal energy. In this the aluminium cube does positive work since it is expanding against atmospheric pressure.
 
Here - W (negative )
Is because the work done by the system during heating ?

and why delta T in C why not in kelvin (K)

thank you
 
Last edited:
The aluminium does positive work in expanding against the atmospheric pressure, that is the surface exerts an outwards force against the air as it expands. So the direction of movement (expansion) is in the same direction as the force it is exerting, so one inserts it as positive work in the formula

ΔU = Q - W

Weh ndifference of temperature are involved you need not change the temperature to kelvin since their is just a constant difference between the two 273 degrees.
 
The system will loose energy if it does work, like the expanding block pushing outwards against the air. Because the force that it is exerting against the air is directed outwards and the surface is also moving in this direction, outwards, so it is doing positive work. If the block was cooled down it would still need to support itself against the atmosphere, so it is still pushing back against it, but now the direction of motion is inwards as it is contracting while it is cooling, but now the work done by it is negative since the force and direction of motion are opposite to each other. Also since the atmosphere is pushing it inwards we get that the block will gain energy because the atmosphere is doing work on it. So positive work means it is losing energy and negative work means it is gaining energy.
 
thanks a lot
 

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