How does temperature in Celcius affect enthalpy in a compressed liquid?

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SUMMARY

The discussion focuses on the relationship between temperature in Celsius and enthalpy in compressed liquids, specifically referencing the equation Cv*T=h. At 50°C, the calculated enthalpy (hf) is 209.34 kJ/kg, aligning closely with the result from the equation using Cv≈4.18. The conversation highlights that the reference state for zero enthalpy is not absolute zero and emphasizes the importance of consulting property tables for accurate reference states. Understanding these concepts is crucial for accurate thermodynamic calculations.

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  • Understanding of thermodynamic properties, specifically enthalpy.
  • Familiarity with specific heat capacity (Cv) and its application in calculations.
  • Knowledge of property tables for compressed liquids and their reference states.
  • Basic grasp of temperature scales, particularly Celsius and Kelvin.
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  • Research the significance of reference states in thermodynamics.
  • Study the properties of saturated water tables and their temperature dependencies.
  • Learn how to accurately use property tables for various fluids.
  • Explore the implications of using different temperature scales in thermodynamic equations.
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Students and professionals in thermodynamics, chemical engineering, and mechanical engineering who are looking to deepen their understanding of enthalpy calculations and the effects of temperature on thermodynamic properties.

MariusM
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Hello!

I was playing around with a problem and while I was doing it I noticed a similarity between

Cv*T=h by using celsius and not Kelvin.

So if I have a compressed liquid at say 50°C, looking up at a property table I find the hf to be 209.34 kJ/kg (the pressure is considered to be low). If you take the equation Cv*T, where Cv≈4.18 and T=50°C you get 209 kJ/kg. Doing this with kelvin instead gives a much larger number.

My question is why is the relation here celsius? Does it have something to do with the saturated water tables being celsius dependent?
 
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MariusM said:
Hello!

I was playing around with a problem and while I was doing it I noticed a similarity between

Cv*T=h by using celsius and not Kelvin.

So if I have a compressed liquid at say 50°C, looking up at a property table I find the hf to be 209.34 kJ/kg (the pressure is considered to be low). If you take the equation Cv*T, where Cv≈4.18 and T=50°C you get 209 kJ/kg. Doing this with kelvin instead gives a much larger number.

My question is why is the relation here celsius? Does it have something to do with the saturated water tables being celsius dependent?
The reference state for zero enthalpy is typically not taken as absolute zero. You have to look up in your table the specific reference state that they use for zero enthalpy. This information is usually given on the first page of the table at the very top. Or, you can look at the numbers in the table until you find a set of conditions where h is zero.

Chet
 
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Thank you very much for your informative reply! I will read more upon the subject of reference state regarding zero enthalpy so I'll hopefully get a full comprehension of this.
 

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