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, 1 cal = 4.184 J ( 1 cal 4 .184 J ) and ( 4 .184 J 1 cal ) One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow: 1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal ) and ( 1000 cal 1 kcal ) 1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ )

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Answer 1

Question

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

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Answer 2

Question

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

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Check out a sample textbook solution

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Answer 3

Question

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

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Answer 4

Question

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

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Answer 5

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

Answer 6

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

Answer 7

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,
1 cal = 4.184 J( 1 cal 4.184 J )and ( 4.184 J 1 cal)

One kilocalorie (kcal) is equal to 1000 calories, and one kilojoule (kJ) is equal to 1000 joules. The equalities and conversion factors follow:
1 kcal = 1000 cal ⇒ ( 1 kcal 1000 cal) and ( 1000 cal 1 kcal)1 kJ = 1000 J ⇒ ( 1 kJ 1000 J ) and ( 1000 J 1 kJ)

Answer 8

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Answer 10

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Answer 11

Solution Summary: The author explains that the energy required to keep a 75 watt light bulb burning for 3.0 h should be calculated.

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