Thermodynamics- How much Ice is melted?

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SUMMARY

The discussion focuses on a thermodynamics problem involving the melting of ice when 0.175 kg of water at 88.0°C is poured into an insulated cup containing 0.212 kg of ice at 0°C. The key equations used include the heat transfer equation Qwater = -Qice and q = mc(Tf - Ti), with specific heat capacities for water (Cwater = 4187 J/kg°C) and ice (Cice = 2090 J/kg°C). The main challenge is determining the mass of ice melted at thermal equilibrium, requiring an understanding of heat flow and the relationship between temperature change and mass change.

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  • Understanding of thermodynamic principles, specifically heat transfer.
  • Familiarity with the specific heat capacities of water and ice.
  • Knowledge of the heat transfer equation Q = mcΔT.
  • Ability to solve algebraic equations involving multiple variables.
NEXT STEPS
  • Study the concept of latent heat and its role in phase changes, particularly for ice melting.
  • Learn how to apply conservation of energy principles in thermodynamic systems.
  • Explore examples of thermal equilibrium problems involving different substances.
  • Investigate the implications of insulated systems in thermodynamics.
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Students studying thermodynamics, educators teaching physics concepts, and anyone interested in understanding heat transfer and phase changes in materials.

seichan
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Homework Statement


0.175 kg of water at 88.0 degC is poured into an insulated cup containing 0.212 kg of ice initially at 0 degC. How many kg of liquid will there be when the system reaches thermal equilibrium?

Homework Equations


Qwater=-Qice
q=mc(Tf-Ti)
Cwater= 4187 J/kg degC
Cice=2090 J/kg degC [not sure on this one, had to look it up]

The Attempt at a Solution


Alright, I know how to get the final temperature of the solution:
q=mc(Tf-Ti)
.175(4187)(Tf-Ti)=-.212(2090)(Tf-Ti)
.175(4187)Tf-.175(4220)(88+273)=-.212(2090)Tf
.175(4187)Tf+.212(2090)Tf=.175(4187)(88)
Tf=[.175(4187)(88)]/[.175(4187)+.212(2090)]
What I'm not sure of is how much of the ice this melts into water... I considered putting the temperature back into the equilibrium equation and solving for how much mass it must take, but I'm very confused as to how to denote the change in mass, considering the fact that the final masses of both the ice and the water are unknown.
Qwater=-Qice
(mi(water)-mf(water))(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)]-88)=-(mi(ice)-mf(ice)(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)])
(.175-mf(water))(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)]-88)=-(.212-mf(ice)(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)])

Any help would be appreciated.
 
Last edited:
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Hi seichan,

You do use q=m c (\Delta T) for the heat flow in or out required to change the temperature of a mass m of a substance. However, here the ice is initially at zero degrees celsius. As the heat initially begins flowing into the ice, it begins melting, but it's temperature does not change until it completely melts. How is heat flow related to the mass of ice that has melted?
 

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