Physics

Power, Energy, and Efficiency

463 Questions

Power, energy, and efficiency are fundamental physics concepts that quantify work and the rate of doing work. Questions cover electrical energy consumption in kilowatt hours, the efficiency of thermodynamic cycles like the Brayton cycle, and mechanical power lifting loads. This topic is essential for general science and engineering exam preparation.

Electrical energy consumptionBrayton cycle efficiencyPower calculation formulasCommercial electricity unitsHydraulic lift mechanics

Power, Energy, and Efficiency Questions

Multiple choice physics work and energy commercial unit of energy power work and power

1 kilowatt-hour is the amount of .... by 1000 watt electric appliance when it operates for one hour.

  1. Power

  2. Voltage

  3. Electric energy

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1 kilowatt-hour is the amount of energy energy consumed by a 1000 watt electric appliance when it is operated for one hour.

Multiple choice physics work and energy commercial unit of energy power work and power

A lamp rated 20w and an electric iron rated 50w are used for 2 hour everyday. Calculate the total energy consumed in 20 days.

  1. 14kwh

  2. 2.8kwh

  3. 40kwh

  4. All
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy consumed by lamp in 2 hour     $E _l = 0.02\times 2 = 0.04$ kWh per day

Energy consumed by iron in 2 hour     $E _i = 0.05\times 2 = 0.1$ kWh per day
$\therefore$ Total energy consumed by both appliance     $E _T = (E _l+E _i)\times 20 = (0.04+0.1)\times 20 = 2.8$ kWh

Multiple choice physics work and energy commercial unit of energy power work and power

One kilowatt hour is equal to

  1. $\displaystyle 36\times { 10 }^{ 5 }$ joules
  2. $\displaystyle 36\times { 10 }^{ 3 }$ joules
  3. $\displaystyle { 10 }^{ 3 }$ joules
  4. $\displaystyle { 10 }^{ 5 }$ joules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1$ kW  $ = 1000$ $\dfrac{J}{s}$            

$1$ h $ = 3600$ s
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =36\times 10^5$  $J$

Multiple choice physics work and energy commercial unit of energy power work and power

Number of KWh in 1Joule.

  1. $\displaystyle 3.6\times { 10 }^{ 6 }KWh$
  2. $\displaystyle 2.77\times { 10 }^{ -7 }KWh$
  3. $\displaystyle 600KWh$
  4. $\displaystyle 1.6\times { 10 }^{ -19 }KWh$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know  $1$ kWh $ = 3.6\times 10^6$ $J$

$\therefore$  $1$ $J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$ kWh

Multiple choice physics work and energy commercial unit of energy power work and power

Calculate the number of Joules in 1KWh.

  1. $\displaystyle 6\times { 10 }^{ -19 }J$
  2. $\displaystyle 3.6\times { 10 }^{ 6 }J$
  3. $\displaystyle 60J$
  4. $\displaystyle 59J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1$ kW $ = 1000$ $\dfrac{J}{s}$

$1$ h $=3600$ s 
$\therefore$  $1$ kWh $ = 1000\dfrac{J}{s}\times 3600$ $s  =3.6\times 10^6$  $J$

Multiple choice physics work and energy commercial unit of energy power work and power

$1kWh= $ _________?

  1. $3600000\ J$
  2. $10000\ J$
  3. $4.2\ J$
  4. $25000\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Kilowatt hour is the energy consumed by a body of power $1\ kW$ in $1\ hr$. 
Hence, 
$1\ kWh = 1 kW \times 1\ hr$
              $=  10^3 W \times 3600\ s$
              $= 3600000\ J$
Multiple choice physics work and power commercial unit of energy power work and energy

A small diesel engine uses a volume of $1.5 \times 10^4\, cm^3$  of fuel per hour to produce a useful power
output of 40 kW. It may be assumed that 34 kJ of energy is transferred to the engine when it uses $1.0\, cm^3$  of fuel.
What is the rate of transfer from the engine of energy that is wasted?

  1. 850 kW

  2. 920 kW

  3. 840 kW

  4. 810 kW
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy produced by $1.0\ cm^3$ of fuel = $34\ kJ$

So, energy produced by  $1.5 \times 10^4\ cm^3$ of fuel in one hour = $1.5 \times 10^4 \times 34\ kJ$
                                                                                                   = $5.1 \times 10^5\ kJ$
Energy produced in one second = $\dfrac{5.1 \times 10^5}{60}\ kJ/s$
                                                         = $850\ kW$
So, rate of energy wasting = $(850-40)\ kW$
                                             = $810\ kW$

Multiple choice physics work and energy commercial unit of energy power work and power

1 kWh is equal to

  1. $3.6 \times 10^6 MJ$
  2. $3.6 \times 10^5 MJ$
  3. $3.6 \times 10^2 MJ$
  4. $3.6 MJ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
1 kilowatt hour is the energy produced by 1 kilowatt  power source in 1 hour.

$1kWh=1kW\times 1hour=1000\times 3600 W.s$

$\implies 1kWh=3.6\times 10^6J$

$\implies 1kWh=3.6MJ$

Answer-(D)
Multiple choice physics work and energy commercial unit of energy power work and power

Number of kilowatt-hours =$\dfrac { volt\times ampere\times time }{ 1000 } $. Then:

  1. time in seconds

  2. time in minutes

  3. time in hours

  4. time in days
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Kilowatt-hours is the power generated in one hour=$\dfrac{volt\times current\times time( hour)}{1000}$


Answer-(C)

Multiple choice physics work and energy commercial unit of energy power work and power

$1$ kWh$=$ ______________J.

  1. $3.6\times 10^6$
  2. $36\times 10^6$
  3. $3.6\times 10^7$
  4. $3.6\times 10^5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1$ kilowatt-hour(kWh) is a unit of energy. Normally, we want energy to be expressed in joules(J) and time in seconds(s).
Energy(kWh)$=$Power(kW)$\times$(h)$=1000$W$\times 3600$s$=1000$J/s$\times 3600$s$=3600000$J$=3.6\times 10^6$J.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

Efficiency of engine is $n _{1}$ at $T _{1}$= $200^\circ C$ and $T _{2}$ = $0^\circ C$ and $n _{2}$ at $T _{1} = 0^\circ C$ and $T _{2}=-200^\circ C$. Find the ratio of $\cfrac {n _{1}}{n _{2}}$

  1. $1.00$
  2. $0.721$
  3. $0.577$
  4. $0.34$4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Efficiency n = 1 - (T2/T1). For n1: T1 = 473K, T2 = 273K, n1 = 1 - (273/473) = 200/473. For n2: T1 = 273K, T2 = 73K, n2 = 1 - (73/273) = 200/273. Ratio n1/n2 = (200/473) * (273/200) = 273/473 approx 0.577.

Multiple choice power work and power work, energy and power physics energy and its forms

The driving side belt has a tension of $1600\ N$ and the slack side has $500\ N$ tension. The belt turns a pulley $40\ cm$ in radius at a rate of $300\ rpm$. The pulley drives a dynamic having $90\%$ efficiency. How many kilowatt are being delivered by the dynamo?

  1. $12.4\ kW$
  2. $6.2\ kW$
  3. $24.8\ kW$
  4. $13.77\ kW$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power = (T1 - T2) * v. v = omega * r = (2 * pi * 300 / 60) * 0.4 = 12.56 m/s. Power = (1600 - 500) * 12.56 = 13816 W = 13.8 kW. Applying 90% efficiency: 13.8 * 0.9 = 12.42 kW.