Question in thermodynamics-melting of ice.

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SUMMARY

The discussion revolves around a thermodynamics problem involving the mixing of 2 kg of ice at -20 degrees Celsius with 5 kg of water at 20 degrees Celsius. The key equations used include the heat transfer equations for both water and ice, specifically the formulas for heat loss and gain, as well as the latent heat of fusion. The initial attempts to find the final temperature resulted in unrealistic values, indicating that not all ice melted. The solution requires calculating the mass of ice that melts and the corresponding heat exchange to determine the final amount of water in the container.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with specific heat capacities: 1 cal/g°C for water and 0.5 cal/g°C for ice.
  • Knowledge of latent heat of fusion, specifically 80 cal/g for ice.
  • Ability to solve algebraic equations involving temperature and mass.
NEXT STEPS
  • Calculate the heat transfer involved in melting ice using the latent heat of fusion.
  • Learn how to set up and solve heat balance equations in thermodynamic systems.
  • Explore the concept of phase changes and their impact on temperature and energy transfer.
  • Study examples of similar thermodynamic problems to reinforce understanding of heat exchange.
USEFUL FOR

Students studying thermodynamics, educators teaching heat transfer concepts, and anyone interested in solving practical physics problems involving phase changes and energy conservation.

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Homework Statement



Ice of mass 2 kg at -20 degrees C and 5 kg water at 20 degrees C are mixed. Find the net amount of water in the container. (Specific heat of water=1cal/g degrees C, specific heat of Ice = 0.5 cal/g degrees C, latent heat of fusion=80cal/g degrees C)

Homework Equations



Heat lost by water=heat gained by ice
Heat=m.C.(t2-t1)
Where m is mass of substance, C is specific heat of substance, t1 is initial temperature, t2 is final temperature.
Heat=m.L
Where, m is mass of substance, and L is latent heat of substance (latent heat of fusion).



The Attempt at a Solution



Well, the first thing I did is to equate the heat lost by water to the heat gained by the ice. Putting all the variables in the equation,

(2kg)(0.5 cal/g degreesC)(T- (-20 degrees C))=(5kg)(1cal/g degrees C)(T-20 degrees C)
Where, I have assumed T to be the final temperature.

On solving the equation, I'm getting T=30 degrees C which is obviously not possible since, it is higher than the temperature of water.

This probably means that the final temperature is higher than zero.
So I add the latent heat equation.
(2kg)(0.5 cal/g degrees C)(0-(-20)) + (2kg)(80cal/g degrees C)+(2kg)(0.5 cal/g degrees C)(T-0)=(5kg)(1cal/g degrees C)(T-20)
T=70 degrees C (Again, not possible)

I'm assuming this means that all the ice has not melted. But, I have no idea as to how to proceed from this point. Since it is asking the amount of water in the container, some amount of ice has melted and some amount has not melted. How do I ascertain how much ice has melted?
Help will be appreciated.
Thanks.
 
Last edited:
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What gets heated= what gets cooled
Ice will get heated from -20 C to ice at 0 C, the ice at 0 C will get heated, changing to water at 0 C = Water getting cooled from 20 C to water at 0 C

I would find out when ice gets heated from -20 to 0 how much the water gets cooled.
Then I would find out what mass of ice is needed to cool the water from the new temperature to 0 C

Subtract from initial amount of ice to get amount of ice left
 
What gets heated= what gets cooled
Ice will get heated from -20 C to ice at 0 C, the ice at 0 C will get heated, changing to water at 0 C = Water getting cooled from 20 C to water at 0 C

I would find out when ice gets heated from -20 to 0 how much the water gets cooled.
Then I would find out what mass of ice is needed to cool the water from the new temperature to 0 C

Subtract from initial amount of ice to get amount of ice left
 

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