# Question 8e7c8

Jun 4, 2016

$\text{39,800 J}$

#### Explanation:

The idea here is that you need to calculate how much heat is required in order to

• convert your sample of ice at $- {35}^{\circ} \text{C}$ to ice at ${0}^{\circ} \text{C}$
• convert the ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$

In order to be able to calculate the heat required in both cases, you need to know the specific heat of ice and the heat of fusion of water

${c}_{\text{ice" = "2.06 J g"^(-1)""^@"C}}^{- 1}$

$\Delta {H}_{\text{fus" = "334.16 J g}}^{- 1}$

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In this case, ice has a specific heat of ${\text{2.06 J g"^(-1)""^@"C}}^{- 1}$, which means that in order to increase the temperature of $\text{1 g}$ of ice by ${1}^{\circ} \text{C}$, you need to provide it with $\text{2.06 J}$ of heat.

Your tool of choice here will be this equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, you have a change in temperature of

$\Delta T = {0}^{\circ} \text{C" - (-35^@"C") = 35^@"C}$

Plug in your values to find

${q}_{1} = 98.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 35color(red)(cancel(color(black)(""^@"C}}}}$

${q}_{1} = \text{7,065.8 J}$

Now, adding this much heat to $\text{98.0 g}$ of ice at $- {35}^{\circ} \text{C}$ will get you ice at ${0}^{\circ} \text{C}$. In order to melt the ice, you must provide it with enough heat to allow it to undergo a solid $\to$ liquid phase change.

The heat of fusions, $\Delta {H}_{\text{fus}}$, tells you how much heat must be added per gram of solid ice at ${0}^{\circ} \text{C}$ in order to convert it to liquid water at ${0}^{\circ} \text{C}$.

Since you have $\text{98.0 g}$ of ice at ${0}^{\circ} \text{C}$, you will need

98.0 color(red)(cancel(color(black)("g"))) * overbrace("334.16 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)(=DeltaH_"fus")) = "32,748 J"

The total amount of heat needed to get your sample from ice at $- {35}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$ will thus be

${q}_{\text{total" = "7,065.8 J" + "32,748 J}}$

${q}_{\text{total" = "39,814 J}}$

I'll leave the answer rounded to three sig figs

"the amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("39,800 J")color(white)(a/a)|)))#