Energy Transferred Equation : Physics Overview
In this blog, we will discuss the most important energy transferred equation,
- E = Pt
We will also explore other energy transfer equations:
- E=QV
- E=IVt
By the end of this blog, you will understand which equation to use in different conditions, learn about common mistakes to avoid in your exams, and explore real-life applications that will enhance your understanding for future studies.
Contents
Chapter 1
What is Energy Transfer?
Chapter 2
E = Pt (Energy = Power x time)
Chapter 3
E = QV (Energy = Charge x Potential Difference)
Chapter 4
E = IVt (Energy = Current x Potential Difference x time)
Chapter 5
How to Choose the Right Energy Transferred Equation and Some Common Mistakes to Avoid
Chapter 1
What is Energy Transfer?
Energy transfer refers to the movement of energy from one place or form to another.
In physics, energy exists in various forms — such as kinetic, potential, thermal, electrical, chemical, and nuclear—and can be transferred between objects or converted from one form to another within a system.
For a detailed explanation of energy transfer and its significance in physics, please refer to our previous blog post: Stores of Energy
Work Done and Energy Transferred Equation
In physics,
- Work done refers to the energy transferred when a force moves an object over a distance.
- Similarly, in electrical contexts, when electric charges move through a potential difference (voltage), energy is transferred.
- Work done and energy transferred are essentially the same in physics.
- Both quantify how much energy is moved or converted during a process
- Also, both are measured in Joules (J).
Chapter 2
Energy Transferred Equation: E = Pt
The equation E=Pt relates the energy transferred to the power and the time over which the power is applied.
where:
- E is the energy transferred or work done (in joules, J)
- P is the power (in watts, W)
- t is the time (in seconds, s)
Understanding Power
Understanding Power:
- Power (P) is the rate at which work is done or energy is transferred.
- It is measured in watts (W), where 1 watt equals 1 joule per second (1 W = 1 J/s).
Finding Power:
Since,
E=P×t
rearranging it gives
P=E/t
Understanding Power:
- Power (P) is the rate at which work is done or energy is transferred.
- It is measured in watts (W), where 1 watt equals 1 joule per second (1 W = 1 J/s).
Finding Power:
Since,
E=P×t
rearranging it gives
P=E/t
Using E=P×t in Various Scenarios
In Electrical Appliances for Calculating Energy Consumption:
For an electrical device with a known power rating operating for a certain time
Solved Example
Question: A 60 W light bulb is used for 5 hours. How much energy does it consume?
Solution:
- Step #1:
Convert Time to Seconds:
t = 5 hours × 3600 s/hour
= 18,000 s
- Step #2:
Calculate Energy:
E = P×t
= 60 W×18,000 s
= 1,080,000 J
In Mechanics for Calculating Power Requires:
Machines like engines and motors perform work over time. Knowing their power output and the time they operate allows you to calculate the total work done
Solved Example
Question: An electric motor lifts a 200 kg load to a height of 10 meters in 8 seconds. Calculate the power of the motor and the energy transferred.
Solution:
- Step #1: Calculate the Work Done (Energy Transferred):
Work done against gravity:
W=m×g×h
- m=200 kg
- g=9.8 m/s2
- h=10 m
W = 200 kg×9.8 m/s2×10 m= 19,600 JW
- Step #2:
Calculate the Power:P=W / t =19,600 J / 8 s = 2,450 W
- The energy transferred (work done) is 19,600 joules.
- The power of the motor is 2,450 watts.
Chapter 3
Energy Transferred Equation: E = QV
Equation: E = QV (Energy = Charge x Potential Difference)
- E: Energy transferred (in joules, J)
- Q: Charge (in coulombs, C)
- V: Voltage (in volts, V)
This equation is used to find Energy transferred when you know the charge moving through a potential difference (voltage).
Solved Example
An electron moves through a potential difference of 200 V. Calculate the energy transferred.
Solution:
- Step #1:
- Identify the known values:
- Charge of an electron, Q=1.6×10−19 C
- Voltage, V=200 V
- Step #2:
E = QV
(1.6×10−19 C)(200 V)=3.2×10−17 J
Answer: The energy transferred is 3.2×10−17 joules
Chapter 4
Energy Transferred Equation: E = IVt
Equation: E = IVt (Energy = Current x Potential difference x time)
- : Energy transferred (in joules, J)
- : Current (in amperes, A)
- : Voltage (in volts, V)
- : Time (in seconds, s)
Relationship Between Equations
This equation combines
P=IV (power = current x voltage) with
E=Pt (energy = power x time),
resulting in E=IVt
Use this equation when you know the current flowing through a component, the voltage across it, and the time for which it operates. It’s useful for calculating energy transfer in circuits where current and voltage are given.
Solved Example
A device operates at a current of 2 A and a voltage of 12 V for 5 minutes. Calculate the energy transferred.
Solution:
- Step #1:
Convert time to seconds:
t = 5 minutes×60 s/min = 300 s
- Step #2:
- Use the equation E=IVt:
E = IVt = (2 A)(12 V)(300 s) = 7,200 J
Chapter 5
How to Choose the Right Energy Transfer Equation and Some Common Mistakes to Avoid
Formula | When Used |
|---|---|
E = P t | - Use when you know the power of a device and the time it operates. |
- Applicable for calculating energy consumption of electrical appliances or mechanical systems over time. | |
E = Q V | - Use when you know the charge moved through a potential difference. |
- Applicable in scenarios involving the movement of charge in electric fields or circuits. | |
E = I V t | - Use when you know the current, voltage, and time. |
- Useful for calculating energy transferred in electrical circuits where current and voltage are known over a period of time. |
Steps for Choosing the Right Energy Transferred Equation
- Identify Known Quantities:
- List all the values provided in the problem (e.g., power, time, charge, current, voltage).
- Determine What You Need to Find:
- Decide which variable you are solving for (e.g., energy transferred).
Common Mistakes
- Unit Conversions
- Time:
- Always convert time to seconds (s) unless units are consistent.
- Charge:
- Ensure charge is in coulombs (C).
- Energy:
- Energy should be in joules (J).
- Misinterpreting Symbols
- Voltage (V) vs. Velocity (v):
- Be careful with uppercase and lowercase letters.
- Current (I) vs. Time (t):
- Don’t confuse current (I) with the number one.
- Voltage (V) vs. Velocity (v):
- Time:
- Determine What You Need to Find:
- Decide which variable you are solving for (e.g., energy transferred).
Practice Questions on Energy Transferred Equation
Below are the detailed solutions to the practice questions on energy transferred equations. Review each step to understand how to apply the formulas effectively.
Practice Questions
Q1: An electric heater has a power rating of 2,000 W and operates for 3 hours. Calculate the total energy transferred by the heater in joules.
Q3: A light bulb draws a current of 0.5 A when connected to a 230 V supply. Calculate the energy transferred if the bulb is left on for 2 hours.
Q4: An appliance uses 540,000 joules of energy when operating at a power of 1,500 W. For how long (in seconds) was the appliance operating?
Q5: Calculate the energy transferred when a current of 3 A flows through a device with a voltage of 12 V for 5 minutes.
Q6: An electron moves through a potential difference of 1,000 V. The charge of an electron is about 1.6×10−19 C, calculate the energy transferred to the electron in joules.
Q7: A battery supplies a current of 2 A to a circuit for 30 minutes. If the total energy transferred is 72,000 J, what is the voltage of the battery?
Q8: A machine operates at a constant power output of 5,000 W. How much energy does it transfer in 10 minutes?
Below are the detailed solutions to the practice questions on energy transferred equation. Review each step to understand how to apply the formulas effectively.
- Solution to Question 1
Question: An electric heater has a power rating of 2,000 W and operates for 3 hours. Calculate the total energy transferred by the heater in joules.
Solution:
Identify the Formula:
E=P×t
- E: Energy transferred (J)
- P: Power (W)
- t: Time (s)
Convert Time to Seconds:
t=3 hours×3600 s/hour=10,800 s
Calculate Energy:
E=2,000 W×10,800 s=21,600,000 J
Answer:
- The total energy transferred by the heater is 21,600,000 joules.
Solution to Question 2
Question: A charge of 4 coulombs moves through a potential difference of 9 volts. How much energy is transferred?
Solution:
Identify the Formula:
E=Q×V
- E: Energy transferred (J)
- Q: Charge (C)
- V: Voltage (V)
Calculate Energy:
E=4 C×9 V=36 J
Answer:
- The energy transferred is 36 joules.
Solution to Question 3
Question: A light bulb draws a current of 0.5 A when connected to a 230 V supply. Calculate the energy transferred if the bulb is left on for 2 hours.
Solution:
Identify the Formula:
E=I×V×t
- E: Energy transferred (J)
- I: Current (A)
- V: Voltage (V)
- t: Time (s)
Convert Time to Seconds:
t=2 hours×3600 s/hour=7,200 s
Calculate Energy:
E=0.5 A×230 V×7,200 s=828,000 J
Answer:
- The energy transferred is 828,000 joules.
Solution to Question 4
Question: An appliance uses 540,000 joules of energy when operating at a power of 1,500 W. For how long (in seconds) was the appliance operating?
Solution:
Identify the Formula:
E=P×t ⟹ t=E/P
- E: Energy transferred (J)
- P: Power (W)
- t: Time (s)
Calculate Time:
t = 540,000 J/1,500 W = 360 s
Answer:
- The appliance was operating for 360 seconds.
Solution to Question 5
Question: Calculate the energy transferred when a current of 3 A flows through a device with a voltage of 12 V for 5 minutes.
Solution:
Identify the Formula:
E=I×V×t
- E: Energy transferred (J)
- I: Current (A)
- V: Voltage (V)
- t: Time (s)
Convert Time to Seconds:
t=5 minutes×60 s/minute=300 s
Calculate Energy:
E= 3 A×12 V×300 s = 10,800 J
Answer:
- The energy transferred is 10,800 joules.
Solution to Question 6
Question: An electron moves through a potential difference of 1,000 V. The charge of an electron is about 1.6×10−19, calculate the energy transferred to the electron in joules.
Solution:
Identify the Formula:
E=Q×V
- E: Energy transferred (J)
- Q: Charge (C)
- V: Voltage (V)
Calculate Energy:
E=1.6×10−19 C×1,000 V=1.6×10 −16 J
Answer:
- The energy transferred to the electron is 1.6×10−16.
Solution to Question 7
Question: A battery supplies a current of 2 A to a circuit for 30 minutes. If the total energy transferred is 72,000 J, what is the voltage of the battery?
Solution:
Identify the Formula:
E=I×V×t ⟹ V=E / I×t
- E: Energy transferred (J)
- I: Current (A)
- V: Voltage (V)
- t: Time (s)
Convert Time to Seconds:
t=30 minutes×60 s/minute=1,800 s
Calculate Voltage:
V = 72,000 J / 2 A×1,800 s
Answer:
- The voltage of the battery is 20 volts.
Solution to Question 8
Question: A machine operates at a constant power output of 5,000 W. Find the amount of energy transferred in 10 minutes?
Solution:
Identify the Formula:
E=P×t
- E: Energy transferred (J)
- P: Power (W)
- t: Time (s)
Convert Time to Seconds:
t=10 minutes×60 s/minute=600 s
Calculate Energy:
E=5,000 W×600 s=3,000,000 J
Answer:
- The machine transfers 3,000,000 joules of energy.
Solution to Question 9
Question: A resistor in a circuit has a voltage drop of 15 V across it and a current of 0.2 A flows through it for 10 seconds. Calculate the energy dissipated by the resistor.
Solution:
Identify the Formula:
E=I×V×t
- E: Energy transferred (J)
- I: Current (A)
- V: Voltage (V)
- t: Time (s)
Calculate Energy:
E=0.2 A×15 V×10 s=30 J
Answer:
- The energy dissipated by the resistor is 30 joules.
Solution to Question 10
Question: An electric car battery stores 21.6 MJ (megajoules) of energy. If the battery operates at a voltage of 400 V and supplies a current of 90 A, how long (in hours) can the car run before the battery is depleted?
Solution:
Convert Energy to Joules:
21.6 MJ=21.6×10^6
Identify the Formula:
E=I×V×t ⟹ t=E / I×V
- E: Energy (J)
- I: Current (A)
- V: Voltage (V)
- t: Time (s)
Calculate Time in Seconds:
t=21.6×10^6 J / (90 A×400 V) =21.6×10^6 / 36,000=600 s
Convert Time to Hours:
t=600 s / 3,600 s/hour=0.1667 hours≈0.17 hours
Answer:
- The car can run for approximately 0.17 hours (or 10 minutes) before the battery is depleted.