91Ó°ÊÓ

Suppose the snow is composed of $50\mathrm{\%}$ice and $50\mathrm{\%}$air, and that it's $2$metres deep. The sun provides around $1000$watts of power, but around localid="1650328444025" $90\mathrm{\%}$of this is reflected by the snow, so the snow absorbs only about 100 watts.

The density of ice melting point is

$$\begin{gathered} \mathrm{ÒÏ}=0.9167\mathrm{g}â‹\dots {\mathrm{cm}}^{−3} \\ =0.9167\frac{\mathrm{g}}{{\mathrm{cm}}^3}×\frac{{10}^6{\mathrm{cm}}^3}{{\mathrm{m}}^3} \end{gathered}$$

$\mathrm{ÒÏ}=9.167×{10}^5\mathrm{g}â‹\dots {\mathrm{m}}^{−3}$

Because snow is made up of 50% ice and 50% air, there is a mass $m$of ice in a block of snow with a surface area of $1{\mathrm{m}}^2$and a depth of $2\mathrm{m}$:

$$\begin{gathered} m=\frac{50}{100}\mathrm{ÒÏ}×V \\ =\frac{1}{2}\left(9.167×{10}^5\right)g×(2×1) \\ =9.167×{10}^5\mathrm{g} \end{gathered}$$

To melt the ice, you'll need the following quantity of heat:

$Q=m×L$

where $L$is the latent heat, and its for melting ice, so:

$$\begin{gathered} Q=9.167×{10}^5×80 \\ =7.333×{10}^7\mathrm{cal} \end{gathered}$$

but $1\mathrm{cal}=4.184\mathrm{J}$is,

$$\begin{gathered} Q=4.184×7.333×{10}^7\mathrm{cal} \\ =3.068×{10}^8\mathrm{J} \end{gathered}$$

The sun generates heat in the form of:

$P=\frac{Q}{tA}$

where $t$is the amount of time that the region $A$must be exposed to the heat $Q$, the time is:

$t=\frac{Q}{PA}$

Substitute $Q=100\text{watts,}A=1{\mathrm{m}}^2\text{and}Q=3.068×{10}^8\mathrm{cal}\text{, so:}$

$\begin{array}{c}t=\frac{3.068×{10}^8}{100×1}\\ \end{array}$

role="math" localid="1650328384083" $\begin{array}{c}=3068000\mathrm{s}\\ \end{array}$

$\begin{array}{c}\\ t=733300\mathrm{s}=35.5\mathrm{days}\end{array}$