Solve Air Heated @Constant Vol: Find 1Q2

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Discussion Overview

The discussion revolves around calculating the heat transfer (1Q2) for air heated from 300K to 350K at constant volume. Participants explore the relevant equations and concepts related to isochoric processes and the determination of mass or moles of air involved in the calculation.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that since the process is isochoric, no work is done, leading to the equation 1Q2 = m(u2-u1) + 1W2 simplifying to 1Q2 = m * cv * (t2 - t1).
  • Another participant suggests using the equation Q = n Cv delta T and notes that the mass of gas is needed to find the number of moles.
  • A participant proposes an expression for 1Q2 as 50 * 0.717 * m, questioning its correctness.
  • There is a clarification about the relationship between moles and mass, with emphasis on using mass in their thermodynamics class instead of moles.
  • Participants discuss the meaning of 1Q2, questioning whether it refers to the change in heat energy input or change in internal energy.

Areas of Agreement / Disagreement

Participants generally agree that the process is isochoric and that no work is done. However, there is disagreement regarding the use of mass versus moles in calculations and the interpretation of 1Q2.

Contextual Notes

There are unresolved aspects regarding the determination of mass and the specific definitions of terms used, such as 1Q2, which may lead to different interpretations of the calculations.

Who May Find This Useful

Students and educators in thermodynamics, particularly those dealing with heat transfer calculations in isochoric processes.

akhmed966
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Homework Statement


Air is heated from 300 to 350K at constant volume, Find 1Q2.


Homework Equations


1Q2 = m(u2-u1)+1W2
u2-u1 = cv(t2-t1)

The Attempt at a Solution


The PV diagram of the process is straight up, so the area under the curve is 0, so the work done is zero.
Cv is .717 from the back of the book

1Q2 = m*.717(350-300)+0

How do I find the mass? I don't think I'm doing it right.
 
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Its a isochoric process.

you are right about no work done.
using Q = n Cv delta T,

what is left is finding how much moles of air is heat.
if not given, just treat it as a variable. more information required. the mass of gas will tell you how many moles there are.
 
That's what I left it as,

1Q2 = 50*.717*m

Would that be right?
 
moles = mass / (molecular weight x avogadro's constant)
avogadro constant = 6.02 x 10^23

molecular weight is something that you can find on the data booklet.

you can't leave the mass just like tt.
the m you wrote is moles not mass.

may i ask you wad you actually mean by 1Q2. delta heat energy input or delta internal energy.
delta means change in.
 
Last edited:
darkdream said:
moles = mass / (molecular weight x avogadro's constant)
avogadro constant = 6.02 x 10^23

molecular weight is something that you can find on the data booklet.

you can't leave the mass just like tt.
the m you wrote is moles not mass.

may i ask you wad you actually mean by 1Q2. delta heat energy input or delta internal energy.
delta means change in.

Final heat, and in our thermodynamics class, we don't use moles (n), we use m for mass.
 

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