Chapter 3.4, Problem 27PP

Interpretation Introduction

Interpretation: The energy required to keep a 75 −watt light bulb burning for 3.0 h ,in joules and in kilocalories, should be calculated.

Concept Introduction:

• One calorie is defined as 4.184 J. The conversion factors can be written as,
$\begin{array}{l}1\text{cal}=4.184\text{J}\\ \left(\frac{\text{1 cal}}{\text{4}\text{.184 J}}\right)\text{and}\left(\frac{\text{4}\text{.184}\text{J}}{\text{1}\text{cal}}\right)\end{array}$

• One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
$\begin{array}{l}1\text{kcal = 1000}\text{cal}\Rightarrow \left(\frac{\text{1}\text{kcal}}{\text{1000}\text{cal}}\right)\text{and}\left(\frac{\text{1000}\text{cal}}{\text{1}\text{kcal}}\right)\\ 1\text{kJ}\text{=}\text{1000}\text{J}\Rightarrow \left(\frac{\text{1}\text{kJ}}{\text{1000}\text{J}}\right)\text{and}\left(\frac{\text{1000}\text{J}}{\text{1}\text{kJ}}\right)\end{array}$