Max Work from 800 ıC Brick Cooling to 25 ıC

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Discussion Overview

The discussion revolves around the potential for recovering work from the cooling of 1-kg blocks of brick, initially at 800 °C, as they cool to ambient air at 25 °C. Participants explore the theoretical framework for calculating maximum work output, considering thermodynamic principles and the specific heat capacity of the brick.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that a reversible process is optimal for work extraction and proposes using a Carnot cycle, although they express uncertainty about the relevance of the air temperature.
  • Another participant questions the appropriateness of using the equation W=m*Cp*ΔT, arguing that it should apply if the air acts as a constant temperature heat sink.
  • A third participant counters that the equation may not be suitable for air or water due to the dependence of Cp on temperature and pressure.
  • A later reply states that while a reversible cycle is valid, the Carnot cycle is not applicable since it operates between two constant temperatures, and emphasizes the need to consider the entropy changes in the cooling process.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of certain equations and the applicability of the Carnot cycle, indicating that multiple competing perspectives remain unresolved.

Contextual Notes

There are limitations regarding the assumptions made about the cooling process, particularly concerning the temperature dependence of specific heat capacities and the nature of the heat exchange with the environment.

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Homework Statement


A young engineer notices in her plant that 1-kg blocks of brick are routinely removed from a 800 ıC oven and are let to cool in air at 25 ıC. Conscious about cost-cutting and efficiency, she wonders whether some work could be recovered from this process. Calculate the maximum amount of work that could be obtained. The CP of brick is 0.9 kJ/kg K. Can you come up with devices that could extract this work?


Homework Equations


Q=ΔH, ΔU=W+Q, η=1-Tl/Th,


The Attempt at a Solution


I know that a reversible process is the best option do so. This is most likely going to be done with a carnot cycle to extract the work. I'm not sure if the T of the air given is going to be any help. The bricks being 1kg can be relevant to how much thermal energy it has. Don't do this with Q=mCpΔT because we used in basic chemistry thermo so he doesn't want us to use that in this class.

 
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It isn't clear to me why W=m*Cp*ΔT isn't appropriate. Assuming the air is a large heat sink of constant temperature the brick will cool to air temperature. That equation should be appropriate.
 
That equation would work for the brick but not air or water since Cp is actually calculated from tabulated values which depend on temperature and pressure.
 
seeveedubyah said:

Homework Statement


A young engineer notices in her plant that 1-kg blocks of brick are routinely removed from a 800 ıC oven and are let to cool in air at 25 ıC. Conscious about cost-cutting and efficiency, she wonders whether some work could be recovered from this process. Calculate the maximum amount of work that could be obtained. The CP of brick is 0.9 kJ/kg K. Can you come up with devices that could extract this work?


Homework Equations


Q=ΔH, ΔU=W+Q, η=1-Tl/Th,


The Attempt at a Solution


I know that a reversible process is the best option do so. This is most likely going to be done with a carnot cycle to extract the work. I'm not sure if the T of the air given is going to be any help. The bricks being 1kg can be relevant to how much thermal energy it has. Don't do this with Q=mCpΔT because we used in basic chemistry thermo so he doesn't want us to use that in this class.

Carnot is out since that cycle operates between two constant temperatures. Your T2 is constant at 25C but your T1 is not.

Reversible cycle, yes. Be guided by the 1st and 2nd laws. While the entropy increase S2 of the cooling medium is just Q2/T, T = 25 + 273, S1 will be an integral over the temp. drop of the brick and will of course be negative.

So Q1 = Q2 + W but S2 + S1 > 0.
 

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