Chapter U3.3, Problem 6E

Interpretation Introduction

Interpretation : Volume of liquid water that results after melting snow with different densities is to be compared.

Concept Introduction : When snow melts it volume decreases but the mass of water formed after melting remains the same. This is because the same water molecules are present in frozen snow and melted snow. Density of snow is less than density of water. Snow has varied densities.

Answer to Problem 6E

The formula to calculate mass of a substance is given by:

  $\begin{array}{l}\text{Mass(m)=Density(D)×Volume(V)}\\ \text{m=D×V}\end{array}$

Snow with density$\text{0}\text{.5g/mL}$will have more volume of liquid water than snow with density $\text{0}\text{.25g/mL}$

Explanation of Solution

Given information:

Densities of snow are given and the volumes of two snow samples are equal.

When snow melts, its volume decreases but the mass of water formed after melting remains the same. This is because the same water molecules are present in frozen snow and melted snow. Density of snow is less than density of water. We can use density equation to convert between mas and volume.

The formula to calculate mass of a substance is given by:

  $\begin{array}{l}\text{Mass(m)=Density(D)×Volume(V)}\\ \text{m=D×V}\end{array}$

Mass of snow can be calculated as follows:

Given:

  $\text{Density of snow1=0}\text{.5g/mL}$

Let volume of snow 1be VmL

  $\text{m(snow)=D(snow)×V(snow)}$

  $\text{m(snow1)=0}\text{.5×V=0}\text{.5Vg}$

Given: $\text{Density of snow2=0}\text{.25g/mL}$

Volume of snow 2 is same as volume of snow 1 = VmL

  $\begin{array}{l}\text{m(snow2)=D(snow2)×V(snow2)}\\ \text{m(snow2)=0}\text{.25×V=0}\text{.25Vg}\end{array}$

Mass of liquid water is equal to mass of snow

  $\begin{array}{l}\text{m(snow1)=m(water1)=0}\text{.5Vg}\\ \text{m(snow2)=m(water2)=0}\text{.5Vg}\end{array}$

We can calculate volume of water in each case as follows:

(Density of water is taken as $\text{1}\text{.0g/mL}$)

  $\text{V(water1)=}\frac{\text{m(water1)}}{\text{D(water1)}}\text{=}\frac{\text{0}\text{.5V}}{\text{1}}\text{=0}\text{.5VmL}$

  $\text{V(water2)=}\frac{\text{m(water2)}}{\text{D(water2)}}\text{=}\frac{\text{0}\text{.25V}}{\text{1}}\text{=0}\text{.25VmL}$

Since $\text{0}\text{.5VmL>0}\text{.25VmL}$

  $\text{V(water1)>V(water2)}$

Hence Snow with density $\text{0}\text{.5g/mL}$will have more volume of liquid water than snow with density $\text{0}\text{.25g/mL}$

Conclusion

Mathematical equations can be used to convert volume of snowand water. To use mathematical equations, we must know the density of snowand density of liquid water.