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Explain, in terms of the definition of power, why energy consumption is sometimes listed in kilowatt-hours rather than joules. What is the relationship between these two energy units?

Short Answer

Expert verified

The energy consumption is listed in kilowatt-hours rather than joules because the unit Joule is too small to use for everyday electrical energy consumption.

Step by step solution

01

Definition of Concepts

Power consumption: Power consumption is defined as the rate at which energy is being consumed.

Mathematically power is given as,

\(P = \frac{E}{t}\)

Here,\(P\)stands for power,\(E\)stands for energy used, and\(t\)stands for time.

The expression for energy consumed is obtained by rearranging equation (1.1),

\(E = Pt\)

In larger unit power is measured in terms of kilowatt\(\left( {{\rm{kW}}} \right)\)and time is measure in hours\(\left( {\rm{h}} \right)\).

Therefore, the larger unit of power is,

\(\begin{aligned}{}E &= \left( {{\rm{kW}}} \right) \times \left( {\rm{h}} \right)\\ &= {\rm{kWh}}\end{aligned}\)

02

Explain the relationship between kilowatt-hour and joule

One kilowatt-hour of energy equals,

\(\begin{aligned}{}E& = 1{\rm{ kWh}}\\ &= 1000{\rm{ W}} \times {\rm{1 h}}\\ &= \left( {1000{\rm{ J}} \cdot {{\rm{s}}^{ - 1}}} \right) \times \left( {1{\rm{ h}}} \right) \times \left( {\frac{{60{\rm{ min}}}}{{1{\rm{ h}}}}} \right) \times \left( {\frac{{60{\rm{ sec}}}}{{1{\rm{ min}}}}} \right)\\ &= 3.6 \times {10^6}{\rm{ J}}\end{aligned}\)

Therefore, one kilowatt-hour of energy consumption equals \(3.6 \times {10^6}{\rm{ J}}\).

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Most popular questions from this chapter

Using energy considerations, calculate the average force a 60.0-kg sprinter exerts backward on the track to accelerate from 2.00 to 8.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

(a) What is the average useful power output of a person who does \begin{aligned}6.00 \times {10^6}{\rm{ J}} \end{aligned}of useful work in 8.00 h?

(b) Working at this rate, how long will it take this person to lift 2000 kgof bricks 1.50 mto a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)

(a) Calculate the force the woman in Figure 7.46 exerts to do a push-up at constant speed, taking all data to be known to three digits.

(b) How much work does she do if her center of mass rises 0.240 m?

(c) What is her useful power output if she does 25 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.

Figure 7.46 Forces involved in doing push-ups. The woman’s weight acts as a force exerted downward on her center of gravity (CG).

Calculate the power output needed for a 950-kg car to climb a 2.00° slope at a constant 30.0 m/s while encountering wind resistance and friction totaling 600 N. Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy.

(a) How long can you rapidly climb stairs\(\left( {116/{\rm{min}}} \right)\)on the\(93.0{\rm{ kcal}}\)of energy in a\(10.0 - {\rm{g}}\)pat of butter?

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