Another problem (Phase Change Heat Calculation)

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SUMMARY

The discussion centers on phase change heat calculations involving the equations q = mcat and mols * hvap. The user calculated the heat required to decrease the temperature from 70 to 56 degrees Celsius, resulting in q = 1195.5 J, and the heat required for phase change from gas to liquid as 29129 J. A confusion arose regarding the sign of the heat when cooling the liquid, where the user initially calculated q = -7047 J due to the temperature change. The correct approach is to consider the heat removed from the gas, leading to a total heat calculation of 37.37 kJ.

PREREQUISITES
  • Understanding of thermodynamic equations, specifically q = mcat and mols * hvap.
  • Knowledge of phase changes and the associated heat transfer concepts.
  • Familiarity with temperature change calculations and their implications on heat energy.
  • Basic understanding of the principles of cooling and heating in thermodynamics.
NEXT STEPS
  • Study the principles of heat transfer in phase changes, focusing on latent heat calculations.
  • Learn about the specific heat capacities of different substances to enhance understanding of q = mcat.
  • Explore detailed examples of thermodynamic calculations involving multiple phase changes.
  • Investigate common mistakes in thermodynamic calculations, particularly regarding temperature changes and sign conventions.
USEFUL FOR

Students in thermodynamics, chemistry, and physics, as well as educators seeking to clarify phase change heat calculations and common misconceptions in energy transfer.

Lori

Homework Statement



upload_2017-12-13_0-57-4.png

Homework Equations


q = mcat
mols*hvap

The Attempt at a Solution



To decrease temp from 70 to 56 degrees (bp), q=mcat = 1195.5
to convert this gas into liquid , must do mols * hvap = 29129 joules
to decrease temp from 56 to 0 thought, i get q= mcat = -7047 J (which is negative- is this wrong?)

it seems like the correct answer is when 7047 is positive after adding these up,. i get 37.37 kj. which seams like the answer (cant see it cause my professor erased it...)

Why wouldn't i subtract the last number?
 

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No idea where you got the minus from - cooling the liquid is no different from cooling the gas (for which you got the correct answer).
 
Borek said:
No idea where you got the minus from - cooling the liquid is no different from cooling the gas (for which you got the correct answer).
from mcat , i got the negative from change in temperature (final mius initial) so (0-56)
 
So, for colling of liquid ΔT = final - initial = 0 - 56 = -56, yes?

And what was the change in temperature for the cooling of gas?
 
Borek said:
So, for colling of liquid ΔT = final - initial = 0 - 56 = -56, yes?

And what was the change in temperature for the cooling of gas?
Oh I see. We are looking for any heat removed from cooling the gas so we should add heat to cool the liquid since it was used to cool the gas as well
 

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