# Question e02ea

Jan 8, 2016

$\text{350 J}$

#### Explanation:

Before doing anything else, make sure that you understand what happens when a solid melts.

The first thing to take note of here is that since melting is a phase change, it must take place at constant temperature.

As you know, pure water melts and freezes at ${0}^{\circ} \text{C}$. This means that a solid $\to$ liquid phase change takes place when you're going from solid at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$.

Notice that your sample of ice starts at a temperature of $- {1.1}^{\circ} \text{C}$. This means that before you can melt it, you must first heat it from solid at $- {1.1}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$.

This means that you will have to know two things

• the specific heat of ice, which tells you how much heat is needed to increase the temperature of $\text{1 g}$ of ice by ${1}^{\circ} \text{C}$

• the enthalpy of fusion for water, which tells you how much heat is needed in order to convert $\text{1 g}$ of ice from solid to liquid at its melting point

You can find these values here

${c}_{\text{ice" = 2.09"J"/("g" ""^@"C")" }}$

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

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

Two equations will come in handy here

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

color(blue)(q = m * DeltaH_"fus")" ", where

$q$ - heat absorbed / lost
$m$ - the mass of the sample
$\Delta {H}_{\text{fus}}$ - the enthalpy of fusion of the substance

So, to get $\text{3.5 g}$ of ice to go from solid at $- {1.1}^{\circ} \text{C}$ to solid at ${0}^{\circ} \text{C}$, you need to provide it with

q_1 = 3.5 color(red)(cancel(color(black)("g"))) * 2.09"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [(0 - (-1.1)]color(red)(cancel(color(black)(""^@"C"))) = "8.05 J"

To get the solid ice from solid at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$, you'll need

q_2 = 3.5 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g"))) = "337.5 J"

The total heat needed to go from ice at $- {1.1}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$ will thus be

${q}_{\text{total}} = {q}_{1} + {q}_{2}$

${q}_{\text{total" = "8.05 J" + "337.5 J}}$

${q}_{\text{total" = "345.55 J}}$

Rounded to two sig figs, the answer will be

q_"total" = color(green)("350 J")

To get the answer in kilojoules, use the conversion factor

$\text{1 kJ" = 10^3"J}$

q_"total" = 350 * 10^(-3)"kJ" = color(green)("0.35 kJ")#