# Question bf98d

Jan 27, 2014

Density is the amount of stuff inside a volume. In our case, our key equation looks like the following:

$\mathrm{de} n s i t y = \frac{m a s s \setminus o f \setminus i c e}{v o l u m e \setminus o f \setminus i c e}$

We are given the $\mathrm{de} n s i t y$ as $0.617 \frac{g}{c} {m}^{3}$. We want to find out the mass. To find the mass, we need to multiply our density by the total volume of ice.

Eq. 1. $\left(\mathrm{de} n s i t y\right) \cdot \left(v o l u m e \setminus o f \setminus i c e\right) = m a s s \setminus o f \setminus i c e$

Thus, we need to follow the volume of ice and then convert everything into the proper units.

Let's find the volume of ice. We are told 82.4%# of Finland is covered in ice. Thus, the actual area of Finland covered in ice is

$\frac{82.4}{100} \cdot 2175000 \setminus k {m}^{2} = 1792200 \setminus k {m}^{2}$

Notice percentages have no units, so our answer of how much area is covered in ice remains in $k {m}^{2}$.

Now that we have the area of ice covering Finland, we can find the volume. Because we are given the average depth of the ice sheet, we can assume the ice sheet looks roughly like a rectangular prism, or The formula for find the volume of a rectangular prism is just $a r e a \cdot h e i g h t$. We know the $a r e a$, and we are given the $h e i g h t$ or depth as $7045 m$.

$V o l u m e \setminus o f \setminus i c e = 1792200 \setminus k {m}^{2} \cdot 7045 m$
Our units are no equivalent, so we'll need to convert meters into kilometers. There are 1000 meters in a kilometer
$V o l u m e \setminus o f \setminus i c e = 1792200 \setminus k {m}^{2} \cdot \left(7045 m \cdot \frac{1 k m}{1000 m}\right)$
$V o l u m e \setminus o f \setminus i c e = 1792200 \setminus k {m}^{2} \cdot 7.045 k m$
$V o l u m e \setminus o f \setminus i c e = 1792200 \setminus k {m}^{2} \cdot 7.045 k m$
$V o l u m e \setminus o f \setminus i c e = 12626049 \setminus k {m}^{3}$

Now that we have the volume of ice, we can get its mass using Eq. 1.

Eq. 1. $\left(\mathrm{de} n s i t y\right) \cdot \left(v o l u m e \setminus o f \setminus i c e\right) = m a s s \setminus o f \setminus i c e$

Eq. 2. $\left(0.617 \frac{g}{c {m}^{3}}\right) \cdot \left(12626049 \setminus k {m}^{3}\right)$

Our current units of $c {m}^{3}$ and $k {m}^{3}$ cannot cancel out because they're not the same. We'll convert $k {m}^{3}$ into $c {m}^{3}$. A single $k m$ is $1000 m$. $1 m$ is in turn $100 c m$.

$\frac{c m}{k m} = \frac{1 k m}{1 k m} \cdot \frac{1000 m}{1 k m} \cdot \frac{100 c m}{1 m}$

There are $100000 c m$ in $1 k m$. To get how many $c {m}^{3}$ are in a single $k {m}^{3}$, we just need to cube that number. So there are $1 x {10}^{15} c {m}^{3}$ in $1 k {m}^{3}$. Let's plug in this value to Eq. 2.

Eq. 3. $\left(0.617 \frac{g}{c {m}^{3}}\right) \cdot \left(12626049 \setminus k {m}^{3}\right) \cdot 1 x {10}^{15} \frac{c {m}^{3}}{k {m}^{3}}$

By plugging in this value we cancel both $k {m}^{3}$ and $c {m}^{3}$, which leaves us with just grams. However, we want the answer in $k g$. We know there are $1000 g$ in $1 k g$, so let's also plug that into Eq. 3.

$\left(0.617 \frac{g}{c {m}^{3}}\right) \cdot \left(12626049 \setminus k {m}^{3}\right) \cdot 1 x {10}^{15} \frac{c {m}^{3}}{k {m}^{3}} \cdot \frac{1 k g}{1000 g}$

That allows us to cancel $g$ and end up with $k g$, which concludes our dimension analysis.

Plugging these values into the calculator should give you the right answer! That's a ton of ice.