Basic Chemistry (5th Edition)
Basic Chemistry (5th Edition)
5th Edition
ISBN: 9780134138046
Author: Karen C. Timberlake
Publisher: PEARSON
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Chapter 3, Problem 3.47FU

(a)

Interpretation Introduction

Interpretation:

The total kilocalories corresponding to each meal needs to be determined.

Concept Introduction:

  • Energy value or nutritional value of food is the amount of energy stored in 1 gram of the food material. It is expressed in terms of kcal/g or kJ/g
  • The energy value varies across the food spectrum.

(a)

Expert Solution
Check Mark

Answer to Problem 3.47FU

Breakfast = 270 kcal

Lunch = 420 kcal

Dinner = 440 kcal

Explanation of Solution

Breakfast:

Food consumed and the corresponding energy value:

1 Banana = 110 kcal

1 cup non-fat milk = 90 kcal

1 egg = 70 kcal

Total energy = 110 + 90 + 70 = 270 kcal

Lunch:

Food consumed and the corresponding energy value:

1 cup carrots = 50 kcal

3 oz grounded beef = 220 kcal

1 medium apple = 60 kcal

1 cup non-fat milk = 90 kcal

Total energy = 50 + 220 + 60 + 90 = 420 kcal

Dinner:

Food consumed and the corresponding energy value:

3 oz skinless chicken = 110 kcal

1 baked potato = 100 kcal

3 oz broccoli = 30 kcal

1 cup non-fat milk = 90 kcal

Total energy = 2(110) + 100 + 30 + 90 = 440 kcal

(b)

Interpretation Introduction

Interpretation:

The total kilocalories corresponding to each day needs to be determined.

Concept Introduction:

  • Energy value or nutritional value of food is the amount of energy stored in 1 gram of the food material. It is expressed in terms of kcal/g or kJ/g.
  • The energy value varies across the food spectrum.

(b)

Expert Solution
Check Mark

Answer to Problem 3.47FU

Total energy per day = 1130 kcal

Explanation of Solution

Total energy for breakfast = 270 kcal

Total energy for lunch = 420 kcal

Total energy for dinner = 440 kcal

Total energy per day = 270 + 420 + 440 = 1130 kcal

(c)

Interpretation Introduction

Interpretation:

Whether Person C loses or gains weight needs to be determined based on the new diet.

Concept Introduction:

  • Energy value or nutritional value of food is the amount of energy stored in 1 gram of the food material. It is expressed in terms of kcal/g or kJ/g
  • The energy value varies across the food spectrum.

(c)

Expert Solution
Check Mark

Answer to Problem 3.47FU

Person C will lose weight.

Explanation of Solution

It is given that Person C consumes 1800 kcal per day.However,based on the new diet, he will consume only 1130 kcal per. Since the new diet has a lower total calorie intake, it is likely that Person C will lose weight over time.

(d)

Interpretation Introduction

Interpretation:

The number of days taken by Person C in order to lose 5.0 lbs needs to be determined.

Concept Introduction:

  • Energy value or nutritional value of food is the amount of energy stored in 1 gram of the food material. It is expressed in terms of kcal/g or kJ/g
  • The energy value varies across the food spectrum.

(d)

Expert Solution
Check Mark

Answer to Problem 3.47FU

Days taken to lose 5.0 lbs = 26 days

Explanation of Solution

Energy lost per day = 1800 -1130 = 670 kcal

  Number of days taken to lose 1.0 lbs = 3500 kcal × 1 day670 kcal=5.22 daysNumber of days taken to lose 5.0 lbs = 5.0 lbs × 5.22 days1.0 lbs= 26 days

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Chapter 3 Solutions

Basic Chemistry (5th Edition)

Ch. 3.2 - Prob. 3.11QAPCh. 3.2 - What type of change, physical or chemical, takes...Ch. 3.2 - Prob. 3.13QAPCh. 3.2 - Describe each property of the element zirconium as...Ch. 3.3 - Prob. 3.15QAPCh. 3.3 - Prob. 3.16QAPCh. 3.3 - Prob. 3.17QAPCh. 3.3 - Prob. 3.18QAPCh. 3.3 - Prob. 3.19QAPCh. 3.3 - Prob. 3.20QAPCh. 3.4 - Discuss the changes in the potential and kinetic...Ch. 3.4 - Prob. 3.22QAPCh. 3.4 - Prob. 3.23QAPCh. 3.4 - Prob. 3.24QAPCh. 3.4 - Prob. 3.25QAPCh. 3.4 - Prob. 3.26QAPCh. 3.4 - Prob. 3.27QAPCh. 3.4 - Prob. 3.28QAPCh. 3.5 - If the same amount of heat is supplied to samples...Ch. 3.5 - Prob. 3.30QAPCh. 3.5 - Prob. 3.31QAPCh. 3.5 - Prob. 3.32QAPCh. 3.5 - Prob. 3.33QAPCh. 3.5 - Use the heat equation to calculate the energy, in...Ch. 3.5 - Calculate the mass, in grams, for each of the...Ch. 3.5 - Prob. 3.36QAPCh. 3.5 - Prob. 3.37QAPCh. 3.5 - Prob. 3.38QAPCh. 3.6 - Calculate the kilocalories for each of the...Ch. 3.6 - Prob. 3.40QAPCh. 3.6 - Using the energy values for foods (see Table 3.7),...Ch. 3.6 - Prob. 3.42QAPCh. 3.6 - Prob. 3.43QAPCh. 3.6 - Prob. 3.44QAPCh. 3.6 - Prob. 3.45QAPCh. 3.6 - Prob. 3.46QAPCh. 3 - Prob. 3.47FUCh. 3 - Prob. 3.48FUCh. 3 - Prob. 3.49UTCCh. 3 - Prob. 3.50UTCCh. 3 - Prob. 3.51UTCCh. 3 - Prob. 3.52UTCCh. 3 - Prob. 3.53UTCCh. 3 - Prob. 3.54UTCCh. 3 - Prob. 3.55UTCCh. 3 - Prob. 3.56UTCCh. 3 - Prob. 3.57UTCCh. 3 - Prob. 3.58UTCCh. 3 - Prob. 3.59UTCCh. 3 - Prob. 3.60UTCCh. 3 - Prob. 3.61AQAPCh. 3 - Prob. 3.62AQAPCh. 3 - Prob. 3.63AQAPCh. 3 - Prob. 3.64AQAPCh. 3 - Prob. 3.65AQAPCh. 3 - Prob. 3.66AQAPCh. 3 - Prob. 3.67AQAPCh. 3 - Prob. 3.68AQAPCh. 3 - Prob. 3.69AQAPCh. 3 - Prob. 3.70AQAPCh. 3 - Prob. 3.71AQAPCh. 3 - Prob. 3.72AQAPCh. 3 - Prob. 3.73AQAPCh. 3 - Prob. 3.74AQAPCh. 3 - 3.83 On a hot day, the bleach sand gets hot but...Ch. 3 - Prob. 3.76AQAPCh. 3 - Prob. 3.77AQAPCh. 3 - Prob. 3.78AQAPCh. 3 - Prob. 3.79AQAPCh. 3 - Prob. 3.80AQAPCh. 3 - Prob. 3.81AQAPCh. 3 - Use the heat equation to calculate the energy, in...Ch. 3 - Prob. 3.83AQAPCh. 3 - Prob. 3.84AQAPCh. 3 - Prob. 3.85AQAPCh. 3 - Prob. 3.86AQAPCh. 3 - If you want to lose 1 lb of “body fat,” which is...Ch. 3 - Prob. 3.88AQAPCh. 3 - Prob. 3.89AQAPCh. 3 - Prob. 3.90AQAPCh. 3 - Prob. 3.91CQCh. 3 - Prob. 3.92CQCh. 3 - Prob. 3.93CQCh. 3 - Prob. 3.94CQCh. 3 - Prob. 3.95CQCh. 3 - Prob. 3.96CQCh. 3 - Prob. 3.97CQCh. 3 - Prob. 3.98CQCh. 3 - Prob. 1CICh. 3 - Prob. 2CICh. 3 - Prob. 3CICh. 3 - Prob. 4CICh. 3 - Prob. 5CICh. 3 - Prob. 6CI
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