Heat/fusion/vaporization question?

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SUMMARY

The discussion focuses on a thermodynamics problem involving the heat transfer between steam and ice to achieve a final water temperature of 50°C. The key equations used include q=lm for latent heat and q=cmΔT for specific heat. The participant initially calculated the heat required to melt ice and the heat released by steam but arrived at an incorrect mass of steam (9g) instead of the correct answer (33g). The critical concept is that the heat lost by the steam must equal the heat gained by the ice, necessitating accurate calculations of latent heat and temperature changes.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically heat transfer.
  • Familiarity with the equations for latent heat (q=lm) and specific heat (q=cmΔT).
  • Knowledge of phase changes, including heat of fusion (333 J) and heat of vaporization (2256 J).
  • Ability to perform calculations involving temperature changes and mass in calorimetry.
NEXT STEPS
  • Review the principles of calorimetry and heat transfer in thermodynamics.
  • Study the calculations for phase changes, focusing on heat of fusion and heat of vaporization.
  • Practice similar thermodynamics problems involving steam and ice to solidify understanding.
  • Learn about energy conservation in closed systems and how it applies to heat exchange.
USEFUL FOR

Students studying thermodynamics, educators teaching calorimetry, and anyone needing to solve heat transfer problems involving phase changes.

pirates01
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Homework Statement


Given 150g of ice at -15oC, how much steam (at 100oC in g) is required to give a resultant of 50oC water?


Homework Equations


q=lm
q=cmT
heat for vap: 2256j
heat for fusion: 333J


The Attempt at a Solution


I got something like
remaining heat: 49950-2256m
change heat with remaining water: 4.19(150+m)T
and that m(100)/(150+m)=50 (error here?)

In the end, i got something like m=9g, which is too little, cause answers says 33g
 
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pirates01 said:
I got something like
remaining heat: 49950-2256m
What exactly is your reasoning for this "remaining heat"? What is 49950 and how did you get it?
 
remaining heat: 49950-2256m

I got this from q=lm
where 49950 is the joules taken to melt ice to liquid, and 2256m (where m is the mass of gas) is the joules given off to turn steam to liquid.
 
When you end up with water at 50 oC, you need to say that the heat lost by the steam is equal to the heat gained by the ice. In other words,

Heat to turn steam to liquid + Heat to reduce T of condensed steam from 100 oC to 50 oC

is equal to

Heat to turn ice to liquid + Heat to raise T of melted ice from 0 oC to 50 oC.
 

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