# 350g of 0.0°C ice cubes are placed in a glass of 700g of room temperature water (23.0°C) in an experiment. Later, the temperature of the water in the glass is measured at 13.0°C. Does all of the ice melt?

Dec 23, 2015

No.

#### Explanation:

Your strategy here will be to determine whether or not the heat given off by the warmer liquid water is enough to ensure that all $\text{350 g}$ of ice melt at ${0}^{\circ} \text{C}$.

More specifically, you need to first determine how much heat is given off when the temperature of $\text{700 g}$ of liquid water decreases from ${23.0}^{\circ} \text{C}$ to ${13.0}^{\circ} \text{C}$.

To do that, use the following equation

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed/lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

The specific heat of liquid water is equal to 4.18"J"/("g" ^@"C"). Plug in your values to get

$q = 700 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (13.0 - 23.0)color(red)(cancel(color(black)(""^@"C}}}}$

$q = - \text{29,260 J}$

The minus sign is used to symbolize heat lost.

Now, phase changes always take place at constant temperature and depend on the given substance's enthalpy of fusion. In water's case, you have

$\Delta {H}_{f} = 334 \text{J"/"g}$

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

This means that in order to melt $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, you need to provided it with $\text{334 J}$ of heat.

Calculate how much heat would be required to melt all of the ice given to you

350 color(red)(cancel(color(black)("g"))) * "334 J"/(1color(red)(cancel(color(black)("g")))) = "116,900 J"

As you can see, you need significantly more heat than the $\text{29,260 J}$ the liquid water gives off in order for all of the ice to melt.

This of course implies that not all of the ice will melt. In fact, you can use the heat given off by the liquid water to determine exactly how many grams of ice would melt.

$\text{29,260" color(red)(cancel(color(black)("J"))) * "1 g"/(334color(red)(cancel(color(black)("J")))) = "87.6 g}$

Out of the $\text{350-g}$ sample of ice, only $\text{87.6 g}$ will melt in the period of time used for the experiment.

You can actually determine how much ice will melt until the mixture reaches thermal equilibrium.

Calculate how much heat is given off when $\text{700 g}$ of liquid water go from ${23.0}^{\circ} \text{C}$ to ${0.0}^{\circ} \text{C}$.

$q = 700 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (0.0 - 23.0)color(red)(cancel(color(black)(""^@"C}}}}$

$q = - \text{67,298 J}$

This much heat will melt

$\text{67,298" color(red)(cancel(color(black)("J"))) * "1 g"/(334color(red)(cancel(color(black)("J")))) = "201.5 g}$

of the initial $\text{350-g}$ sample of ice.

You can conclude that at thermal equilibrium, the mixture will contain

${m}_{\text{ice" = "350 g" - "201.5 g" = "148.5 g}}$

of ice at ${0}^{\circ} \text{C}$ and

${m}_{\text{water" = "201.5 g" + "700 g" = "901.5 g}}$

of liquid water at ${0}^{\circ} \text{C}$.