Question

In an insulated vessel, $25g$ of ice at $0^{\circ }C$ is added to $600g$ of water at ${18.0}^{\circ }C$.

1. What is the final temperature of the system?
2. How much ice remains when the system reaches equilibrium?

Answer 1

1. If all $250g$ of ice is melted it must absorb energy
$Q_f=m{L}_f=\left(0.250kg\right)(3.33\times {10}^{5}J/kg)=83.3kJ$
The energy released when 600\:g of water cools from ${18.0}^{\circ }C$ to $0^{\circ }C$ is
$|Q|=|mc\Delta T|=\left(0.600kg\right)\left(4186J/k{g}^{\circ }C\right)\left({18.0}^{\circ }C\right)=45.2kJ$
Since the energy required to melt $250g$ of ice at $0^{\circ }C$ exceeds the energy released by cooling $600g$ of water from ${18.0}^{\circ }C$ to $0^{\circ }C$, not all the ice melts and the final temperature of the system (water + ice) must be $0^{\circ }C$.
2. The originally warmer water will cool all the way to $0^{\circ }C$, so it loses $45.2kJ$ to ice. This energy lost by the water will melt a mass of ice $m$, where $Q=m{L}_f$.
Solving for the mass $m=\frac{Q}{L_f}=\frac{45.2\times {10}^{3}J}{3.33\times {10}^{5}J/kg}=0.136kg$
Therefore, the ice remaining $m^{\prime }=0.250kg-0.136kg=0.114kg$
Step by step reasoning is essential for solving a problem like this.