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A watt is a measure of power (the rate of energy change) equal to \(1 \mathrm{~J} / \mathrm{s}\). (a) Calculate the number of joules in a kilowatt-hour. (b) An adult person radiates heat to the surroundings at about the same rate as a 100 -watt electric incandescent lightbulb. What is the total amount of energy in kcal radiated to the surroundings by an adult in 24 hours?

Short Answer

Expert verified
(a) There are 3,600,000 Joules in a kilowatt-hour. (b) An adult person radiates approximately 2063 kilocalories of energy to the surroundings in 24 hours.

Step by step solution

01

Part (a): Calculating Joules in a Kilowatt-hour

First, we need to know the conversion rate from kilowatts to watts and from hours to seconds. We know that: 1 kilowatt (kW) = 1000 watts (W), and 1 hour = 60 minutes = 3600 seconds (s) Given that 1 watt is equal to \(1 \mathrm{~J/s}\), we can calculate the number of joules in a kilowatt-hour by multiplying the power (in watts) by the time (in seconds). Let's convert 1 kilowatt-hour (kWh) to joules: 1 kWh = 1000 W * 3600 s Now, we can calculate the energy in joules by multiplying the power and time: Energy (in joules) = Power (in watts) × Time (in seconds)
02

Part (b): Calculating Energy Radiated by an Adult in 24 Hours

We are given that an adult person radiates heat at a rate of 100 watts, which is equal to \(100 \mathrm{~J/s}\). We need to find the total amount of energy radiated in 24 hours in kcal. First, let's find the total energy in joules: Energy (in joules) = Power (in watts) × Time (in seconds) Time = 24 hours = 24 * 3600 seconds Now, we will convert the energy in joules to energy in kilocalories (kcal). We know that: 1 kcal = 4184 J Energy (in kcal) = Energy (in joules) / 4184 J/kcal Putting it all together, we can find the total energy radiated by an adult in 24 hours in kcal. Now let's calculate the values for both parts of the problem.
03

Calculating the result for Part (a)

We already have the equation for energy in terms of power and time: Energy (in joules) = Power (in watts) × Time (in seconds) Substituting the values for 1 kWh: Energy = 1000 W * 3600 s = 3,600,000 Joules So, there are 3,600,000 Joules in a kilowatt-hour.
04

Calculating the result for Part (b)

First, we'll calculate the total energy in joules in 24 hours: Energy (in joules) = Power (in watts) × Time (in seconds) Substituting the values for an adult person radiating heat at a rate of 100 watts and a time period of 24 hours: Energy = 100 W * (24 * 3600 s) = 8,640,000 Joules Now, we'll convert this energy to kilocalories: Energy (in kcal) = Energy (in joules) / 4184 J/kcal Energy = 8,640,000 J / 4184 J/kcal ≈ 2063 kcal Therefore, an adult person radiates approximately 2063 kilocalories of energy to the surroundings in 24 hours.

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Key Concepts

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

Joules to Kilowatt-hour
Understanding the conversion between Joules and Kilowatt-hours is essential when dealing with energy-related calculations. A Joule (\(J\)) is the unit of energy derived from the International System of Units (SI) and is equivalent to one watt-second. However, in everyday use, larger units like Kilowatt-hours (\(kWh\)) are often used. To convert Joules to Kilowatt-hours, you need to understand the following:
  • 1 Kilowatt = 1000 Watts.
  • 1 hour = 3600 seconds.
  • 1 Kilowatt-hour (kWh) is the energy consumed by a device with a power of 1000 Watts running for one hour.

So, to find out how many Joules are in a Kilowatt-hour, multiply the power in kilowatts by the time in seconds:\[1 \text{kWh} = 1000 \text{W} \times 3600 \text{s} = 3,600,000 \text{J}\]These conversions are critical in energy management, everyday utility calculations, and reducing energy consumption.
Power Measurement
Power measurement is essential to understand energy consumption and usage efficiencies. In science, power is defined as the rate at which energy is transferred or converted. It is measured in Watts (\(W\)), where one watt is equivalent to one joule of energy per second. Understanding power measurement includes:
  • A Watt reflects the power rate and signifies how quickly work is done or heat is transferred.
  • Joules per second is a standardized way of representing power.
  • 100 Watts, a standard light bulb power consumption, can serve as a relatable comparison for various scenarios.

It's crucial to grasp power measurement concepts not only for academic purposes but also for practical energy efficiency applications in daily life. Daily power use, monitored through devices, affects consumption bills significantly. Through power measurement, individuals can better manage the energy spent by household appliances.
Energy in Kilocalories
Energy is often discussed in various units, and kilocalories (kcal) is a commonly used unit, especially in food energy consumption. Calories express the amount of energy your body gets from food and drinks. In scientific contexts, where Joules might be initially used, it's necessary to convert from Joules to Kilocalories.
  • The conversion factor between Joules and kilocalories is: 1 kcal = 4184 Joules.
  • This conversion is key when studying energy metabolism or assessing daily energy requirements. Applying this knowledge helps in tackling questions about energy radiated by entities, such as comparing human heat emission in kcal.

For example, an adult radiating 100 watts over 24 hours releases energy equivalent to 2063 kcal, a key point linking real-world human activity to energy concepts.
Heat Radiation
Heat radiation is the process through which heat is emitted or absorbed in the form of electromagnetic waves. This concept is crucial in understanding how energy is transferred from one body to another without requiring a medium. In everyday scenarios, humans radiate heat as one of their energy balance processes, maintaining body temperature through principles of thermodynamics.
  • Radiation occurs through the emission of infrared radiation.
  • This is the same way the sun heats the earth, and your body naturally emits heat.
  • The rate of heat radiation from the human body can be compared to a 100-watt bulb which provides a concrete idea of the power involved in such processes.

Understanding heat radiation is not just about academic knowledge; it aids in evaluating how our bodies interact with our environment and how efficiently we maintain homeostasis in fluctuating temperatures.

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Most popular questions from this chapter

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) For a given process at constant pressure, \(\Delta H\) is negative. Is the process endothermic or exothermic?

Given the data $$ \begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)-\cdots+2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{~kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \cdots-\rightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{~kJ} \\ 2 \mathrm{~N}_{2} \mathrm{O}(g)-\cdots 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{~kJ} \end{aligned} $$ use Hess's law to calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \stackrel{-\cdots} 3 \mathrm{NO}(g) $$

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)-\longrightarrow \mathrm{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for production of \(0.200 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(2.50 \mathrm{~g}\) of \(\mathrm{AgCl}\). (c) Calculate \(\Delta \mathrm{H}\) when \(0.150 \mathrm{mmol}\) of AgCl dissolves in water.

Calculate \(\Delta E\), and determine whether the process is endothermic or exothermic for the following cases: (a) A system absorbs \(105 \mathrm{~kJ}\) of heat from its surroundings while doing \(29 \mathrm{~kJ}\) of work on the surroundings; (b) \(q=1.50 \mathrm{~kJ}\) and \(w=-657 \mathrm{~J} ;\) (c) the system releases \(57.5 \mathrm{~kJ}\) of heat while doing \(22.5 \mathrm{~kJ}\) of work on the surroundings.

Consider the decomposition of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\), to gaseous acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})\) : $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \quad \Delta H=+630 \mathrm{~kJ} $$ (a) What is the enthalpy change for the reverse reaction? (b) What is \(\Delta H\) for the formation of \(1 \mathrm{~mol}\) of acetylene? (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction? (d) If \(C_{6} \mathrm{H}_{6}(g)\) were consumed instead of \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\), would you expect the magnitude of \(\Delta H\) to increase, decrease, or stay the same? Explain.

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