Calculating Phase Transitions in a Water-Ice Mixture

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Homework Help Overview

The problem involves calculating the amount of ice that must melt to lower the temperature of a given volume of water from 25°C to 0°C. The context is centered around phase transitions and the specific heat capacities of water and ice.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of specific heat capacity versus latent heat in the context of phase changes. There is confusion regarding the correct equations to use, particularly in relation to latent heat and temperature changes. Some participants suggest starting with simpler assumptions to build understanding.

Discussion Status

The discussion is active, with participants exploring different approaches to the problem. Some have proposed equations involving both specific heat and latent heat, while others are questioning the assumptions made in the original setup. There is a recognition of the need to clarify the role of latent heat in the calculations.

Contextual Notes

Participants note the specific heat capacities and latent heat values relevant to the problem, but there is an acknowledgment of confusion regarding how to apply these values correctly in the equations. The discussion reflects a mix of attempts and clarifications without reaching a definitive solution.

liz_p88
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Homework Statement



Ice at 0°C is mixed with 5 x 10^2 mL of H2O at 25°C. How much ice must melt to lower the H2O temp to 0°C?

Homework Equations



mcΔt = -[mcΔt]

The Attempt at a Solution



(500ml)(4.186 kJ/kg*K)(-25) = m(2.1 kJ/kg*K)(25)


996 ml or gram equivalents

I know this is wrong but I can't figure out how to do it
 
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Why use the specific heat capacity of ice? The ice isn't changing temperature, it's undergoing a phase change. Think latent heat.
 
Thats where I get confused. Okay so because of latent heat, I would have to use the equation...mcΔt + mL + mcΔt = -[mcΔt]...or Q=mcΔt + mL? This is where I get confused. The only thing is, I don't know how to plug in the variables. I know it takes 333.7 kJ to change 1 kg of ice to water at 0 celsius
 
liz_p88 said:
I know it takes 333.7 kJ to change 1 kg of ice to water at 0 celsius

Maybe this is the best place to start reasoning, rather than monkeying around with the equations. Assuming 1 g of ice melts, what would be the temperature change of 500 g of water? Can you continue from there?
 
Ok I think I have it. Q gained by the ice = Q lost by the water
mLf + mc(Tf - 0) = mc(Ti - Tf)

m(333.7) + mc(0-0) = (500)(4.186)(25)

= 157 g
 
Nice.
 

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