Ice cube into water/thermal equilibrium

[farmerburns]

Homework Statement

A 400 cm3 glass is filled with 100 g of ice at 0°C and 170 g of water at 25°C.

Characterize the content of the glass after equilibrium has been reached. Neglect heat transfer to and from the environment.

_ g (mass of water)
_ g (mass of ice)
_°C (equilibrium temperature)

Homework Equations

(dQ)=mc(dT)
specific heat of water= 1.00 (cal/gK)
latent heat of fusion for ice= 79.6 (cal/g)
specific heat of ice= 0.5 (cal/gK)?

The Attempt at a Solution

I have tried this as many ways as I can think of. First off, is volume even an issue? Second, do I need to use heat of fusion to find the final temp.? I feel very lost.

 

[rl.bhat]

m(w)*C*(T2-T1) = m(i)*L + m(i)*T1
The volume of the glass is given to make shure that water will not spill out of the glass when ice and water is filled in the glass.

 

[farmerburns]

m(w)*C*(T2-T1) = m(i)*L + m(i)*T1
The volume of the glass is given to make shure that water will not spill out of the glass when ice and water is filled in the glass.

Using the given equation I get T2=71.8C. Seems a bit high to me. Any other suggestions? I used intial temp. of water=25C, and initial temp of ice=0C.

 

[rl.bhat]

Amount of heat needed to melt the ice = 100*79.6 cal.
Heat lost by the water to reach 0 degree C is = 175*25 cal, which is less than the above. So the ice does not melt completely. Final temperature of the mixture is zero.

 

[farmerburns]

yes, you are right, thank you. I figured it out today with some help.