Thermodynamics-Finding Internal energy

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SUMMARY

The discussion focuses on calculating internal energy and work done during a Carnot cycle involving 0.5 kg of air with an efficiency (η) of 0.5. The highest temperature (Th) is determined to be 585.4 K, and the lowest temperature (Tl) is 292.7 K. The work done during the process W23 is calculated as 107.5 kJ using the formula W23 = m(u2 – u3). The participants seek clarification on how to find internal energies u2 and u3, as well as the work done during the isothermal process.

PREREQUISITES
  • Carnot cycle principles
  • Thermodynamic tables for air properties
  • Internal energy calculations
  • Isothermal process work equations
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  • Study the derivation of internal energy equations for ideal gases
  • Learn how to use thermodynamic tables effectively
  • Research the work done in isothermal processes in detail
  • Explore the efficiency calculations for Carnot cycles
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tre2k3
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I am going over this example

0.5 kg of air undergoes a Carnot cycle with η = 0.5.
Given the initial pressure p1 = 700 kPa, initial
volume V1 = 0.12 m3 and heat transfer during the
isothermal expansion process Q12 = 40 kJ, Find
the highest and the lowest temperatures in the
cycle.
(b) the amount of heat rejection.
(c) work in each process.

They have Th to be 585.4k and Tl to be 292.7k
and so
W23 = m(u2 – u3)= 0.5(423.7 – 208.8) = 107.5 kJ

I do not know how to findu2 and u3. Looking at the tables at he back of the book doesn't help because they don't list the temperatures in this problem. So what equation is used to find u2 and u3 with the given information.
 
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What is the work done during an isothermal process?

What is the efficiency of a carnot cycle?
 

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