Chapter 11, Problem 69A

Interpretation Introduction

Interpretation: The mass of the ice in kilograms should be calculated which measures $25.0\text{ cm }\times \text{42}\text{.0 cm }\times \text{38}\text{.0 cm}$ and has a density $\text{0}{\text{.917 g/cm}}^3$ .

Concept Introduction: Density is defined as the ratio of the mass per unit volume of a substance.

Answer to Problem 69A

The mass of the ice is $\text{36}\text{.6 kg}$ .

Explanation of Solution

The formula for density is:

  $\text{density = }\frac{\text{mass}}{\text{volume}}$  ....... (1)

The volume of the ice cube is:

  $25.0\text{ cm }\times \text{42}\text{.0 cm }\times \text{38}{\text{.0 cm = 39900 cm}}^3$

Substituting the values of density and volume of the ice cube in equation (1) to calculate the mass of the ice cube:

  $\begin{array}{c}\text{0}{\text{.917 g/cm}}^{\text{3}}\text{ = }\frac{\text{mass}}{{\text{39900 cm}}^{\text{3}}}\\ {\text{mass = 39900 cm}}^{\text{3}}\text{×0}{\text{.917 g/cm}}^{\text{3}}\\ \text{mass = 36588}\text{.3 g}\end{array}$

Converting the obtained mass of the ice cube from g to kg using the following conversion factor:

  $\text{1 kg = 1000 g}$

So,

  $\text{36588}\text{.3 g }\times \frac{\text{1 kg}}{1000\text{ g}}\text{ = 36}\text{.5883 kg }\simeq \text{ 36}\text{.6 kg}$