# Thermodynamics- How much Ice is melted?

1. Apr 30, 2008

### seichan

1. The problem statement, all variables and given/known data
0.175 kg of water at 88.0 degC is poured into an insulated cup containing 0.212 kg of ice initially at 0 degC. How many kg of liquid will there be when the system reaches thermal equilibrium?

2. Relevant equations
Qwater=-Qice
q=mc(Tf-Ti)
Cwater= 4187 J/kg degC
Cice=2090 J/kg degC [not sure on this one, had to look it up]

3. The attempt at a solution
Alright, I know how to get the final temperature of the solution:
q=mc(Tf-Ti)
.175(4187)(Tf-Ti)=-.212(2090)(Tf-Ti)
.175(4187)Tf-.175(4220)(88+273)=-.212(2090)Tf
.175(4187)Tf+.212(2090)Tf=.175(4187)(88)
Tf=[.175(4187)(88)]/[.175(4187)+.212(2090)]
What I'm not sure of is how much of the ice this melts into water... I considered putting the temperature back into the equilibrium equation and solving for how much mass it must take, but I'm very confused as to how to denote the change in mass, considering the fact that the final masses of both the ice and the water are unknown.
Qwater=-Qice
(mi(water)-mf(water))(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)]-88)=-(mi(ice)-mf(ice)(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)])
(.175-mf(water))(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)]-88)=-(.212-mf(ice)(4187)([.175(4187)(88)]/[.175(4187)+.212(2090)])

Any help would be appreciated.

Last edited: Apr 30, 2008
2. Apr 30, 2008

### alphysicist

Hi seichan,

You do use $q=m c (\Delta T)$ for the heat flow in or out required to change the temperature of a mass m of a substance. However, here the ice is initially at zero degrees celsius. As the heat initially begins flowing into the ice, it begins melting, but it's temperature does not change until it completely melts. How is heat flow related to the mass of ice that has melted?