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The energy unit used by electrical utilities in their monthly bills is the kilowatt hour \((\mathrm{kWh} ; 1\) kilowatt used for 1 hour). Calculate how many joules there are in a kilowatt hour. If electricity costs \(\$ .09\) per kilowatt hour, calculate how much it costs per megajoule.

Short Answer

Expert verified
1 kWh = 3.6 million joules; cost per megajoule = 2.5 cents.

Step by step solution

01

Understand the Conversion from Kilowatts to Watts

The unit kilowatt hour (kWh) is defined as 1 kilowatt (kW) of power being used continuously over 1 hour. Since 1 kW = 1000 watts (W), we convert kilowatts to watts: 1 kWh = 1000 W for 1 hour.
02

Convert Hours to Seconds

1 hour consists of 3600 seconds (60 minutes * 60 seconds). So we need to express the duration in seconds for the energy calculation.
03

Calculate Energy in Joules per Kilowatt Hour

Since power in watts is energy per unit time in seconds (1 Watt = 1 Joule/second), we calculate the total energy:\[ 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ seconds} = 3,600,000 \text{ Joules} \]
04

Calculate Cost per Kilowatt Hour in Joules

Given that 1 kWh costs $0.09, we first calculate how much 1 joule costs by dividing the cost by the energy amount:\[ \text{Cost per joule} = \frac{0.09}{3,600,000} \approx 2.5 \times 10^{-8} \text{ dollars per joule} \]
05

Convert Cost to Megajoules

1 megajoule (MJ) equals 1,000,000 joules. Multiply the cost per joule by 1,000,000 to get the cost per megajoule:\[ \text{Cost per MJ} = 2.5 \times 10^{-8} \times 1,000,000 = 0.025 \text{ dollars or 2.5 cents} \]
06

Verify Units and Conclusion

Verify that the units all correspond correctly through each step, ultimately finding that the cost per megajoule is 2.5 cents. This confirms the conversions were accurate.

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Key Concepts

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

Electrical Energy Calculation
Understanding electrical energy calculation begins with knowing the unit of power, which is the watt. The energy consumed by a device is measured in kilowatt-hours (kWh). One kilowatt-hour is the amount of energy used when a device consuming one kilowatt (1000 watts) operates for one hour.

Here's a simple way to remember the conversion steps:
  • 1 kilowatt = 1000 watts
  • 1 hour = 3600 seconds
So, when performing an electrical energy calculation, multiply the power in watts by the time in seconds to derive the energy in joules (since 1 watt = 1 joule/second). Understanding these relationships makes it easier to convert and calculate energy use in various forms.
Joules to Kilowatt Hours
Transitioning from joules to kilowatt-hours involves understanding their relationship in terms of time and power. Since 1 kWh represents using 1000 watts over an hour, converting this to joules allows students to grasp larger energy scales.

Here's how you can compute the conversion:
  • 1 kWh = 1000 watts for 3600 seconds (which is equal to one hour)
  • Calculate the energy in joules: 1000 watts * 3600 seconds = 3,600,000 joules
So, each kilowatt-hour equals 3.6 million joules. This conversion is crucial for understanding energy consumption and costs effectively, as small units like joules add up significantly over time.
Cost Calculation of Electricity
Calculating the cost of electricity involves translating the amount of energy consumed, measured in kilowatt-hours or joules, into monetary terms. Since electricity bills often cite consumption in kWh, knowing its equivalent in joules helps in more detailed energy assessments.

Let's break down the cost calculation:
  • First, determine the cost per kilowatt-hour (e.g., $0.09 per kWh).
  • Convert this cost to a per joule basis by dividing the cost by the number of joules in a kWh (3,600,000 joules).
  • Finally, convert to megajoules, where 1 megajoule equals 1,000,000 joules, to understand larger scale costs.
For example, at $0.09 per kWh, the cost per megajoule is approximately 2.5 cents. This method uncovers a detailed look at how electrical energy consumption influences bills, empowering consumers to manage their electricity usage and expenses better.

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Most popular questions from this chapter

$$ \begin{aligned} &\text { Use these bond enthalpy values to answer Question } { . }\\\ &\begin{array}{lclc} \hline \text { Bond } & \begin{array}{c} \text { Bond Enthalpy } \\ (\mathrm{k}\rfloor / \mathrm{mol}) \end{array} & \text { Bond } & \begin{array}{c} \text { Bond Enthalpy } \\ (\mathrm{k} / / \mathrm{mol}) \end{array} \\ \hline \mathrm{H}-\mathrm{F} & 566 & \mathrm{~F}-\mathrm{F} & 158 \\ \mathrm{H}-\mathrm{Cl} & 431 & \mathrm{Cl}-\mathrm{Cl} & 242 \\ \mathrm{H}-\mathrm{Br} & 366 & \mathrm{Br}-\mathrm{Br} & 193 \\ \mathrm{H}-\mathrm{I} & 299 & \mathrm{I}-\mathrm{I} & 151 \\ \mathrm{H}-\mathrm{H} & 436 & & \\ \hline \end{array} \end{aligned} $$ Which molecule, HF, HCl, HBr, or HI, has the strongest chemical bond?

Calorimetric measurements show that the reaction of magnesium with chlorine releases \(26.4 \mathrm{~kJ}\) per gram of magnesium reacted. (a) Write a balanced chemical equation for the reaction. (b) Calculate the standard formation enthalpy of magnesium chloride.

Based on your experience, when ice melts to liquid water, is the process exothermic or endothermic? When liquid water freezes to ice at \(0{ }^{\circ} \mathrm{C}\), is this exothermic or endothermic? (Assume that the ice/water is the system in each case.) Explain your answers.

We burn 3.47 g lithium in excess oxygen at constant atmospheric pressure to form \(\mathrm{Li}_{2} \mathrm{O} .\) Then, we bring the reaction mixture back to \(25^{\circ} \mathrm{C}\). In this process \(146 \mathrm{~kJ}\) of heat is given off. Calculate the standard formation enthalpy of \(\mathrm{Li}_{2} \mathrm{O}\)

Calculate the quantity of energy, in joules, required to raise the temperature of \(454 \mathrm{~g}\) tin from room temperature, \(25.0^{\circ} \mathrm{C}\), to its melting point, \(231.9{ }^{\circ} \mathrm{C},\) and then melt the tin at that temperature. (The specific heat capacity of tin is \(0.227 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{~K}^{-1}\), and the enthalpy of fusion of this metal is \(59.2 \mathrm{~J} / \mathrm{g} .\) )

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