Q. 1.48 - An Introduction to Thermal Physics

Question

 When spring finally arrives in the mountains, the snow pack may be two meters deep, composed $50\mathrm{\%}$ of ice and $50\mathrm{\%}$ of  air. Direct sunlight provides about $1000{\text{ watts/m }}^2$ to earth's surface, but the snow might reflect $90\mathrm{\%}$ of this energy. Estimate how many weeks the snow pack should last, if direct solar radiation is the only source of energy.

Step-by-Step Solution

Answer

 $t=733300\mathrm{s}=35.5\text{ days }$ the anow pack last.

Step :1 Snow composed

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 $90\mathrm{\%}$ of this is reflected by the snow, so the snow absorbs only about 100 watts.

Step : 2 Density of melting point

The density of ice melting point is 

$$\begin{gathered} \rho =0.9167\mathrm{g}\cdot {\mathrm{cm}}^{-3} \\ =0.9167\frac{\mathrm{g}}{{\mathrm{cm}}^3}\times \frac{{10}^6{\mathrm{cm}}^3}{{\mathrm{m}}^3} \end{gathered}$$

$\rho =9.167\times {10}^5\mathrm{g}\cdot {\mathrm{m}}^{-3}$

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 $2\mathrm{m}$:

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

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

$Q=m\times L$

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

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

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

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

Step : 3 Sun generates heat

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

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

$\begin{array}{c}=3068000\mathrm{s}\\ \end{array}$

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