Calculating Phase Transitions in a Water-Ice Mixture

AI Thread Summary
To determine how much ice must melt to cool 500 mL of water from 25°C to 0°C, the latent heat of fusion is crucial. The equation Q gained by the ice equals Q lost by the water is used, incorporating both the latent heat and the specific heat capacities. The calculation shows that melting approximately 157 grams of ice will achieve the desired temperature change. The confusion arises from the need to consider phase changes rather than just temperature changes. This approach effectively balances the heat exchange between the melting ice and the cooling water.
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 celcius
 
liz_p88 said:
I know it takes 333.7 kJ to change 1 kg of ice to water at 0 celcius

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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