Thermodynamics adding ice to water problem with latent heat

[MattNguyen]

Homework Statement

Initially you have mW = 4.6 kg of water at TW = 74°C in an insulated container. You add ice at TI = -19°C to the container and the mix reaches a final, equilibrium temperature of Tf = 33°C. The specific heats of ice and water are cI = 2.10×103 J/(kg⋅°C) and cW = 4.19×103 J/(kg⋅°C), respectively, and the latent heat of fusion for water is Lf = 3.34×105 J/kg.

Enter an expression for the mass of ice you added, in terms of the defined quantities.

Homework Equations

Q=mL
Q=mcT

The Attempt at a Solution

I set this up and then solved for MI: mW(TW-Tf)CW - mI(Tf-TI)CI = (mW + MI)L
The program said that there were additional terms in the numerator and denominator, so I don't know what I'm doing wrong.

mI =​

( mW ( TW - Tf ) cW - mW Lf )/( Lf + ( Tf - TI ) cI )|​

​

 

[Andrew Mason]

Homework Statement

Initially you have mW = 4.6 kg of water at TW = 74°C in an insulated container. You add ice at TI = -19°C to the container and the mix reaches a final, equilibrium temperature of Tf = 33°C. The specific heats of ice and water are cI = 2.10×103 J/(kg⋅°C) and cW = 4.19×103 J/(kg⋅°C), respectively, and the latent heat of fusion for water is Lf = 3.34×105 J/kg.

Enter an expression for the mass of ice you added, in terms of the defined quantities.

Homework Equations

Q=mL
Q=mcT

The Attempt at a Solution

I set this up and then solved for MI: mW(TW-Tf)CW - mI(Tf-TI)CI = (mW + MI)L
The program said that there were additional terms in the numerator and denominator, so I don't know what I'm doing wrong.

mI = ​

( mW ( TW - Tf ) cW - mW Lf )/( Lf + ( Tf - TI ) cI )|​

Hi Matt.

The heat flow from the water changes the initial mass of ice in three stages, all of which have different heat capacities. You need to use the heat capacity of ice in bringing it to 0C then the latent heat of ice to melt it, and finally the heat capacity of liquid water when bringing it from 0C to Tf. They are all different.

AM