Calculating Phase Transitions in a Water-Ice Mixture

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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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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