# Question 6033f

Dec 19, 2017

Here's how you can do that.

#### Explanation:

For starters, you know that in order to melt $\text{100 g}$ of ice at its normal melting point, i.e. to go from ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need

100 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "33,355 J"

This is the case because the enthalpy of fusion of water, $\Delta {H}_{\text{fus}}$, which tells you the energy needed to convert $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, is equal to ${\text{333.55 J g}}^{- 1}$.

This tells you that in order to convert $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid awter at ${0}^{\circ} \text{C}$, you need to provide $\text{333.55 J}$ of heat.

Now, you only have $\text{2,000 J}$ of energy available, so right from the start, you can say for a fact that not all the ice will be converted to liquid water at ${0}^{\circ} \text{C}$.

overbrace("33,355 J")^(color(blue)("what you need")) " " > " " overbrace("2,000 J")^(color(blue)("what you have"))

To find the mass of ice that would melt, use the enthalpy of fusion again.

2000 color(red)(cancel(color(black)("J"))) * "1 g"/(333.55color(red)(cancel(color(black)("J")))) = "5.996 g"

This means that only

$\text{5.996 g " ~~ " 6 g}$

of ice will melt, the rest will remain as ice, not as liquid water. The final temperature of the ice + liquid water mixture will be ${0}^{\circ} \text{C}$ because you didn't provide enough heat to

1. Melt all the ice to get liquid water at ${0}^{\circ} \text{C}$
2. Warm the liquid water to a temperature that is higher than ${0}^{\circ} \text{C}$

So you only converted $\text{6 g}$ of ice at ${0}^{\circ} \text{C}$ to $\text{6 g}$ of liquid water at ${0}^{\circ} \text{C}$, i.e. you performed a solid $\to$ liquid phase change, while leaving $\text{94 g}$ of ice at ${0}^{\circ} \text{C}$.

overbrace("100 g ice at 0"^@"C")^(color(blue)("ice at 0"^@"C"))" " stackrel(color(white)(acolor(red)("+ 2,000 J")aaa))(->) overbrace(" 6 g ")^(color(blue)("liquid water at 0"^@"C")) + overbrace(" 94 g ")^(color(blue)("ice at 0"^@"C"))#