Heat Transfer Coefficient for Phase Change

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SUMMARY

The discussion focuses on calculating the time required for a 100-gram sphere of ice at 0°C to melt in one liter of water at 30°C, utilizing the heat transfer coefficient formula. The initial heat transfer coefficient for ice is established as 0.523 cal/sec/M²°C, with an initial surface area of approximately 0.01034 m² and a temperature difference of 30°C. The calculated heat transfer rate is 0.162 cal/sec, leading to an estimated melting time of 13.7 hours for the ice, although this does not account for the decreasing area and temperature difference during the melting process. The heat of fusion for ice is confirmed to be 80 cal/g, necessitating a reevaluation of the approach used in the calculations.

PREREQUISITES
  • Understanding of heat transfer principles, specifically the heat transfer coefficient.
  • Familiarity with calorimetry and the concept of heat of fusion.
  • Basic knowledge of thermodynamics, particularly temperature gradients.
  • Ability to perform calculations involving surface area and heat transfer rates.
NEXT STEPS
  • Research the application of the heat transfer coefficient in phase change scenarios.
  • Study the effects of varying surface area and temperature on heat transfer rates.
  • Explore advanced calorimetry techniques for more accurate melting time predictions.
  • Learn about numerical methods for solving heat transfer problems involving changing conditions.
USEFUL FOR

Students in thermodynamics, engineers involved in heat transfer applications, and researchers studying phase change phenomena will benefit from this discussion.

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Homework Statement


One liter of water at 30 C (30000 calories )
A 100 gram sphere of ice at 0 C in center of water volume .
The ice will absorb 80000 calories melting, and final water temperature
= 22000 cal/1100g = 20 C.
Assume mixing and uniform water temp during melting and water vessel
insulated from surroundings. How long for ice to melt ?

Homework Equations


Heat Transfer Coefficient
h =q/A delta T
q = cal/sec
h = cal/sec/M2 C for ice , .523 cal/sec ( converted watts to cal/sec )
A = surface area. for approx 100 cc ice = .01034 M2 ( initial)
delta T 30 C ( initial )

The Attempt at a Solution


This is difficult since both Area and delta T are decreasing
For initial conditions only: q = ( .523 cal/sec/ M2 C ) ( .01034 M2(30 C
q= .162 cal/sec. so 8000 cal/ .162cal/sec. = 13.7 hrs
But in the real case this is not correct as A and delta T are decreasing
 
Last edited:
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morrobay said:

Homework Statement


One liter of water at 30 C (30000 calories )
A 100 gram sphere of ice at 0 C in center of water volume .
The ice will absorb 80000 calories melting, and final water temperature
= 22000 cal/1100g = 20 C.
Assume mixing and uniform water temp during melting and water vessel
insulated from surroundings. How long for ice to melt ?

Homework Equations


Heat Transfer Coefficient
h =q/A delta T
q = cal/sec
h = cal/sec/M2 C for ice , .523 cal/sec ( converted watts to cal/sec )
A = surface area. for approx 100 cc ice = .01034 M2 ( initial)
delta T 30 C ( initial )



The Attempt at a Solution


This is difficult since both Area and delta T are decreasing
For initial conditions only: q = ( .523 cal/sec/ M2 C ) ( .01034 M2(30 C
q= .162 cal/sec. so 8000 cal/ .162cal/sec. = 13.7 hrs
But in the real case this is not correct as A and delta T are decreasing

1. That should be 8000 calories melting: ice heat of fusion (80 cal/g) (100g)
Can anyone do this problem or should something else be applied in place of
Heat Transfer Coefficient ?
 

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