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The nutritional calorie (Calorie) is equivalent to 1 kcal. One pound of body fat is equivalent to about \(4.1 \times 10^{3}\) Calories. Express this quantity of energy in joules and kilojoules.

Short Answer

Expert verified
The energy equivalent to one pound of body fat is \(1.71544 \times 10^7\) joules or \(1.71544 \times 10^4\) kilojoules.

Step by step solution

01

Understand the given data

One pound of body fat is equivalent to about \(4.1 \times 10^3\) Calories. These Calories need to be converted into joules and kilojoules.
02

Convert Calories to kilocalories

Since 1 Calorie is equivalent to 1 kilocalorie (kcal), \(4.1 \times 10^3\) Calories is equal to \(4.1 \times 10^3\) kilocalories (kcal).
03

Convert kilocalories to joules

Use the conversion factor: 1 kilocalorie (kcal) is equivalent to 4184 joules (J). Multiply \(4.1 \times 10^3\) kilocalories by 4184 to get the energy in joules. \(4.1 \times 10^3\) kcal \(\times 4184\) J/kcal = \(1.71544 \times 10^7\) J.
04

Convert joules to kilojoules

Since 1 kilojoule (kJ) is equivalent to 1000 joules (J), convert the obtained value in joules to kilojoules by dividing by 1000. \(1.71544 \times 10^7\) J \(\times \frac{1}{1000}\) = \(1.71544 \times 10^4\) kJ.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

nutritional calorie
When talking about energy in nutrition, we often hear the term 'Calorie'. A nutritional Calorie, often capitalized to distinguish it from the smaller 'calorie', is used to measure the energy content of food. One nutritional Calorie is equivalent to one kilocalorie (kcal). This means that if a food item has 100 Calories, it actually contains 100 kilocalories. Understanding this difference is key to managing dietary intake effectively.
kcal to joules conversion
Energy conversion is a fundamental concept in chemistry and nutrition. To convert kilocalories (kcal) to joules (J), we use the fact that 1 kcal is equal to 4184 J. This conversion factor is crucial because it allows us to express energy in different units. For example, if you have 4.1 x 10^3 kcal, you would multiply by 4184 to get the energy in joules. Therefore, 4.1 x 10^3 kcal times 4184 J/kcal equals approximately 1.71544 x 10^7 J. Converting units like this helps scientists and nutritionists compare energy values across different contexts.
energy in kilojoules
Sometimes, we need to convert large amounts of energy into more manageable units. After converting kilocalories to joules, the next step is often to convert joules to kilojoules. One kilojoule (kJ) equals 1000 joules (J). By dividing the energy value in joules by 1000, we can convert it to kilojoules. From our previous calculation, 1.71544 x 10^7 J equals 1.71544 x 10^4 kJ. This simplification can make it easier to grasp and communicate large energy values.
pound of body fat energy
A pound of body fat contains a significant amount of stored energy. Specifically, one pound of body fat is approximately equivalent to 4.1 x 10^3 Calories. This energy can be converted into more universally used units like joules and kilojoules for scientific comparison. Knowing the energy content of body fat can be useful for those studying metabolism, weight management, and energy expenditure. Just remember, 4.1 x 10^3 Calories convert to about 1.71544 x 10^7 J or 1.71544 x 10^4 kJ.
unit conversion in chemistry
Unit conversion is a vital skill in chemistry and other sciences. Whether you're converting mass, volume, or energy, knowing how to switch between different units accurately is important. Converting nutritional Calories to joules and kilojoules is just one example. By mastering these conversions, you can better understand and communicate measurements in scientific contexts. Use conversion factors like 1 kcal = 4184 J and 1 kJ = 1000 J to simplify these calculations. Always double-check your steps to ensure accuracy.

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Most popular questions from this chapter

Acetylene burns in air according to the following equation: $$\begin{array}{r}\mathrm{C}_{2} \mathrm{H}_{2}(g)+{ }_{2}^{5} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H_{\mathrm{ru}}^{\circ}=-1255.8 \mathrm{~kJ}\end{array}$$Given \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{CO}_{2}(g)=-393.5 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{H}_{2} \mathrm{O}(g)=$$-241.8 \mathrm{~kJ} / \mathrm{mol},\) find \(\Delta H_{i}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\)

Why do heat \((q)\) and work \((w)\) have positive values when entering a system and negative values when leaving?

Compounds of boron and hydrogen are remarkable for their unusual bonding (described in Section 14.5 ) and for their reactivity. With the more reactive halogens, for example, diborane \(\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)\) forms trihalides even at low temperatures: $$\begin{array}{r}\mathrm{B}_{2} \mathrm{H}_{6}(g)+6 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{BCl}_{3}(g)+6 \mathrm{HCl}(g) \\\\\Delta H=-755.4 \mathrm{~kJ}\end{array}$$ What is \(\Delta H\) per kilogram of diborane that reacts?

The calorie \((4.184 \mathrm{~J})\) is defined as the quantity of energy needed to raise the temperature of \(1.00 \mathrm{~g}\) of liquid water by \(1.00^{\circ} \mathrm{C}\). The British thermal unit (Btu) is defined as the quantity of energy needed to raise the temperature of 1.00 lb of liquid water by \(1.00^{\circ} \mathrm{F}\) (a) How many joules are in \(1.00 \mathrm{Btu}\left(1 \mathrm{lb}=453.6 \mathrm{~g} ; 1.0^{\circ} \mathrm{C}=1.8^{\circ} \mathrm{F}\right) ?\) (b) The therm is a unit of energy consumption and is defined as 100,000 Btu. How many joules are in 1.00 therm? (c) How many moles of methane must be burned to give 1.00 therm of energy? (Assume that water forms as a gas.) (d) If natural gas costs \(\$ 0.66\) per therm, what is the cost per mole of methane? (Assume that natural gas is pure methane.) (e) How much would it cost to warm 318 gal of water in a hot tub from \(15.0^{\circ} \mathrm{C}\) to \(42.0^{\circ} \mathrm{C}\) by burning methane \((1 \mathrm{gal}=3.78 \mathrm{~L}) ?\)

Diamond and graphite are two crystalline forms of carbon. At 1 atm and \(25^{\circ} \mathrm{C},\) diamond changes to graphite so slowly that the enthalpy change of the process must be obtained indirectly. Using equations from the list below, determine \(\Delta H\) for C(diamond) \(\longrightarrow\) C (graphite) (1) C(diamond) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-395.4 \mathrm{~kJ}\) (2) \(2 \mathrm{CO}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)\) \(\Delta H=566.0 \mathrm{~kJ}\) (3) C(graphite) \(+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H=-393.5 \mathrm{~kJ}\) (4) \(2 \mathrm{CO}(g) \longrightarrow \mathrm{C}\) ( graphite) \(+\mathrm{CO}_{2}(g) \quad \Delta H=-172.5 \mathrm{~kJ}\)

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