In an insulated vessel, `25g` of ice at `0^(@)C` is added to `600g` of water at `18.0^(@)C`.a. What is the final temperature of the system? B. How much ice remains when the system reaches equilibrium?

Explanation

a. If all `250g` of ice is method it must absorb energy
`Q=mL_(f)=(0.250kg)(3.33xx10^(-5)J//kg)=83.3kJ`
The energy released when `600g` of water cools from `18.0^(@)C` to `0^(@)C` is
`|Q|=|mcDeltaT|=(0.600kg)(4186J//kg^(@)C)(18.0^(@)C)=45.2KJ`
Since the energy required to melt `250g` of ice at `0^(@)C` exceeds the energy released by cooling `600g` of water from `18.0^(@)C` not all the ice melts and the final temperature of the system (water+ice) must be `0^(@)C`.
b. The originally warmer water will cool all the way to `0^(@)C` so it loses `45.2KJ` to ice. This energy lost by the water will melt a mass of ice m, where `Q=mL_(f)`.
Solving for the mass `m=(Q)/(L_(f))=(45.2xx10^(3)J)/(3.33xx10^(5)J//Kg)=0.136kg`
Therefore, the ice remaining is `m'=0.250kg-0.136kg=0.114kg`
Step by step reasoning is essential for solving a problem like this.