A person obtains approximately \(2.5 \times 10^{3}\) Calories a day from his or her food. How much energy is that in a. calories? b. joules? c. kilowatt-hours?

Short Answer

Expert verified
a. \(2.5 \times 10^{6}\) caloriesb. \(1.046 \times 10^{7}\) joulesc. \(2.906 \times 10^{-3}\) kilowatt-hours

Step by step solution

01

Understanding Calorie to calorie conversion

Recognize that the term 'Calorie' used in the context of food is actually a kilocalorie. Therefore, to convert food Calories to calories, multiply by 1,000 since 1 food Calorie is equal to 1,000 calories.
02

Converting food Calories to calories

To convert from food Calories to calories, multiply the number of food Calories by 1,000. Calculation: \(2.5 \times 10^{3}\) food Calories \(\times 1,000 = 2.5 \times 10^{6}\) calories.
03

Understanding calorie to joule conversion

1 calorie is defined as 4.184 joules. To convert calories to joules, multiply the number of calories by 4.184.
04

Converting calories to joules

To find the energy in joules, multiply the number of calories by 4.184 joules/calorie. Calculation: \(2.5 \times 10^{6}\) calories \(\times 4.184\) joules/calorie \(= 1.046 \times 10^{7}\) joules.
05

Understanding joule to kilowatt-hour conversion

1 kilowatt-hour is equivalent to 3.6 \(\times\) 10^6 joules. To convert joules to kilowatt-hours, divide the number of joules by 3.6 \(\times\) 10^6.
06

Converting joules to kilowatt-hours

To convert the energy in joules to kilowatt-hours, divide the energy in joules by 3.6 \(\times\) 10^6 joules/kilowatt-hour. Calculation: \(1.046 \times 10^{7}\) joules \(/ 3.6 \times 10^{6}\) joules/kilowatt-hour \(\approx 2.906 \times 10^{-3}\) kilowatt-hours.

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Key Concepts

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

Calorie to calorie conversion
In discussing food energy, it is vital to distinguish between the small calorie, often used in chemistry, and the large calorie, commonly referred to as the food Calorie, denoted with a capital 'C'. The food Calorie is equivalent to a kilocalorie, meaning it is 1,000 times larger than the small calorie. Consequently, when converting food Calories to calories, one needs to multiply by 1,000. For instance, if a person consumes approximately 2,500 food Calories a day, the energy intake in small calories is actually 2,500,000. This difference between the calorie scales is crucial for accurate nutritional and scientific calculations.

Understanding this concept simplifies the complexity of food energy content displayed on nutrition labels, thereby allowing one to better manage their dietary intake. It is also an essential standard in scientific studies where precise measurement of energy is required.
calorie to joule conversion
When converting energy from calories to joules, the standard conversion factor that is utilized is 1 calorie equals 4.184 joules. This factor is rooted in the definition of the calorie as the amount of heat needed to raise the temperature of one gram of water by one degree Celsius. Therefore, to find out how much energy is in joules, we multiply the calories by 4.184. For example, if we start with 2,500,000 calories (as calculated from the previous conversion of food Calories), and we want to express this energy in joules, we multiply it by 4.184 to get approximately 10,460,000 joules.

  • This conversion is fundamental in fields such as nutrition, physics, and any other that demands the precise measurement of energy transfer or consumption.
  • Moreover, understanding this can help in grasping the basic principles of thermodynamics and energy efficiency in biological systems.
joule to kilowatt-hour conversion
The conversion from joules to kilowatt-hours is particularly useful when considering energy usage in electrical appliances or even at a larger scale in power generation and consumption. A kilowatt-hour (kWh) is a unit commonly used in the electrical industry and represents the energy consumption of one thousand watts running for one hour. The conversion factor here is that one kilowatt-hour equals 3.6 million joules (or 3.6 x 10^6 joules).

To perform this conversion, we divide the energy in joules by the factor of 3.6 million. Taking our previous example of 10,460,000 joules, and converting this to kilowatt-hours gives us approximately 0.002906 kWh.

  • This understanding enables students to better appreciate the scale of energy consumption and production, and the efficiency of various energy sources and appliances.
  • It promotes energy literacy which is an invaluable skill in today's world driven by sustainability and energy conservation efforts.

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Most popular questions from this chapter

The amount of \(\mathrm{CO}_{2}\) in the atmosphere is \(0.04 \%(0.04 \%\) \(=0.0004 \mathrm{~L} \mathrm{CO}_{2} / \mathrm{L}\) atmosphere). The world uses the equivalent of approximately \(4.0 \times 10^{12} \mathrm{~kg}\) of petroleum per year to meet its energy needs. Determine how long it would take to double the amount of \(\mathrm{CO}_{2}\) in the atmosphere due to the combustion of petroleum. Follow each of the steps outlined to accomplish this: a. We need to know how much \(\mathrm{CO}_{2}\) is produced by the combustion of \(4.0 \times 10^{12} \mathrm{~kg}\) of petroleum. Assume that this petroleum is in the form of octane and is combusted according to the following balanced reaction: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{~L})+25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{~g})+18 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ By assuming that \(\mathrm{O}_{2}\) is in excess, determine how many moles of \(\mathrm{CO}_{2}\) are produced by the combustion of \(4.0 \times 10^{12} \mathrm{~kg}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\). This will be the amount of \(\mathrm{CO}_{2}\) produced each year. b. By knowing that \(1 \mathrm{~mol}\) of gas occupies \(22.4 \mathrm{~L}\), determine the volume occupied by the number of moles of \(\mathrm{CO}_{2}\) gas that you just calculated. This will be the volume of \(\mathrm{CO}_{2}\) produced per year. c. The volume of \(\mathrm{CO}_{2}\) presently in our atmosphere is approximately \(1.5 \times 10^{18} \mathrm{~L}\). By assuming that all \(\mathrm{CO}_{2}\) produced by the combustion of petroleum stays in our atmosphere, how many years will it take to double the amount of \(\mathrm{CO}_{2}\) currently present in the atmosphere from just petroleum combustion?

A chocolate chip cookie contains about \(200 \mathrm{kcal}\). How many kilowatt- hours of energy does it contain? How long could you light a 100-W light bulb with the energy from the cookie?

Define each of the following terms: a. heat b. energy c. work d. system e. surroundings f. exothermic reaction g. endothermic reaction h. enthalpy of reaction i. kinetic energy j. potential energy

Calculate the amount of carbon dioxide (in \(\mathrm{kg}\) ) emitted into the atmosphere by the complete combustion of a 15.0-gallon tank of gasoline. Do this by following these steps: a. Assume that gasoline is composed of octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\). Write a balanced equation for the combustion of octane. b. Determine the number of moles of octane contained in a 15.0-gallon tank of gasoline (1 gallon = \(3.78\) L). Octane has a density of \(0.79 \mathrm{~g} / \mathrm{mL}\). c. Use the balanced equation to convert from moles of octane to moles of carbon dioxide, then convert to grams of carbon dioxide, and finally to \(\mathrm{kg}\) of carbon dioxide.

The useful energy that comes out of an energy transfer process is related to the efficiency of the process by the following equation: $$ \begin{aligned} &\text { total } \\ &\text { consumed } \end{aligned} \times \text { efficiency }=\begin{aligned} &\text { useful } \\ &\text { energy } \end{aligned} $$ where the efficiency is in decimal (not percent) form. a. If a process is \(30 \%\) efficient, how much useful energy can be derived if \(455 \mathrm{~kJ}\) are consumed? b. A person eats approximately \(2200 \mathrm{kcal} /\) day. How much of that energy is available to do physical work? c. If a car needs \(5.0 \times 10^{3} \mathrm{~kJ}\) to go a particular distance, how much energy will be consumed if the car is \(20 \%\) efficient? d. If an electrical power plant produces \(1.0 \times 10^{9} \mathrm{~J}\) of electrical energy, how much energy will be consumed by the plant if it is \(34 \%\) efficient?

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