# Symbolic algebra

ⓘ Hint: This calculator supports symbolic math. You can enter numbers, but also symbols like a, b, pi or even whole math expressions such as (a+b)/2. If you still don't sure how to make your life easier using symbolic algebra check out our another page: Symbolic calculations

# Inputs data - value and unit, which we're going to convert#

Value | ||

Unit | ||

Decimals |

# $1$ (joule) is equal to:#

# common use#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

joule | Show source$J$ | J | Show source$\text{...}$ | - | The basic energy unit in the SI system. One joule corresponds to the work done by a force of one newton (1 N) by shifting the point of force application by one meter (1 m) in a direction parallel to the direction of the force.$1\ J = 1\ N \cdot 1\ m$ | Show source$...$ |

calorie | Show source$cal$ | cal | Show source$\text{...}$ | - | Heat unit standardized during Fifth International Conference on Water and Steam Properties (IAPWS), which took place in 1956 in London. Unit is an attempt to organize various calorie definitions by introducing so-called international calorie corresponding to exactly 4.868 of joule. See other calorie definitions to learn more.$1\ cal_{IT} \equiv 4.1868\ J$ | Show source$...$ |

kilo-calorie | Show source$kcal$ | kcal | Show source$\text{...}$ | - | Equivalent to one thousand of international calories (1000 cal_{IT}). Sometimes called also large calorie. See the calorie unit for more.$1\ kcal = 1000\ cal_{IT}$ | Show source$...$ |

kilowatt-hour | Show source$kW \times h$ | kW·h | Show source$\text{...}$ | - | Unit used for electric cost pricing. One kilowatt-hour (1 kW·h) corresponds to the amount of electric energy consumed (or approximately emited in the form of heat) by a device with power of one kilowatt (1 kW) within one hour (60 min).$1\ kW \cdot h = 1000\ W \cdot 60\ min = 1000\ \frac{J}{\cancel{s}} \cdot 3600\ \cancel{s} = 3.6\ MJ$ | Show source$...$ |

# SI#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

yottajoule | Show source$YJ$ | YJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One yottajoule is equal to septylion of joules: $1\ YJ= 10^{24}\ J$ | Show source$...$ |

zettajoule | Show source$ZJ$ | ZJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One zettajoule is equal to sextillion of joules: $1\ ZJ= 10^{21}\ J$ | Show source$...$ |

exajoule | Show source$EJ$ | EJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One exajoule is equal to quintillion of joules: $1\ EJ= 10^{18}\ J$ | Show source$...$ |

petajoule | Show source$PJ$ | PJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One petajoule is equal to quadrillion of joules: $1\ PJ= 10^{15}\ J$ | Show source$...$ |

terajoule | Show source$TJ$ | TJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One terajoule is equal to trillion of joules: $1\ TJ= 10^{12}\ J$ | Show source$...$ |

gigajoule | Show source$GJ$ | GJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One gigajoule is equal to billion of joules: $1\ GJ= 10^{9}\ J$ | Show source$...$ |

megajoule | Show source$MJ$ | MJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One megajoule is equal to million of joules: $1\ MJ=1000000\ J= 10^{6}\ J$ | Show source$...$ |

kilojoule | Show source$kJ$ | kJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One kilojoule is equal to thausand of joules: $1\ kJ=1000\ J= 10^{3}\ J$ | Show source$...$ |

hektojoule | Show source$hJ$ | hJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One hektojoule is equal to hundred of joules: $1\ hJ=100\ J= 10^{2}\ J$ | Show source$...$ |

joule | Show source$J$ | J | Show source$\text{...}$ | - | The basic energy unit in the SI system. One joule corresponds to the work done by a force of one newton (1 N) by shifting the point of force application by one meter (1 m) in a direction parallel to the direction of the force.$1\ J = 1\ N \cdot 1\ m$ | Show source$...$ |

decijoule | Show source$dJ$ | dJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One decijoule is equal to one tenth of joule: $1\ dJ=0.1\ J= 10^{-1}\ J$ | Show source$...$ |

centijoule | Show source$cJ$ | cJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One centijoule is equal to one hundredth of joule: $1\ cJ=0.01\ J= 10^{-2}\ J$ | Show source$...$ |

milijoule | Show source$mJ$ | mJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One milijoule is equal to one thousandth of joule: $1\ mJ=0.001\ J= 10^{-3}\ J$ | Show source$...$ |

microjoule | Show source$\mu J$ | µJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One microjoule is equal to one millionth of joule: $1\ \mu J=0.000001\ J= 10^{-6}\ J$ | Show source$...$ |

nanojoule | Show source$nJ$ | nJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One nanojoule is equal to one billionth of joule: $1\ nJ= 10^{-9}\ J$ | Show source$...$ |

pikojoule | Show source$pJ$ | pJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One pikojoule is equal to one trillionth of joule: $1\ pJ= 10^{-12}\ J$ | Show source$...$ |

femtojoule | Show source$fJ$ | fJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One femtojoule is equal to one quadrillionth of joule: $1\ fJ= 10^{-15}\ J$ | Show source$...$ |

attojoule | Show source$aJ$ | aJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One attojoule is equal to one quintillionth of joule: $1\ aJ= 10^{-18}\ J$ | Show source$...$ |

zeptojoule | Show source$zJ$ | zJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One zeptojoule is equal to one sextillionth of joule: $1\ zJ= 10^{-21}\ J$ | Show source$...$ |

yoctojoule | Show source$yJ$ | yJ | Show source$\text{...}$ | - | Derived energy unit in SI system. One yoctojoule is equal to one septillionth of joule: $1\ yJ= 10^{-24}\ J$ | Show source$...$ |

# British Thermal Unit (BTU) related#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

British thermal unit (thermochemical) | Show source$BTU_{th}$ | BTU_{th} | Show source$\text{...}$ | - | An attempt to fix issue with various definitions of heat units. One thermodynamic British thermal unit (1 BTU_{TH}) corresponds exactly to 1.05587 kilojoules. See the other BTU definitions for more information.$1\ BTU_{TH} \equiv 1.05587\ kJ$ | Show source$...$ |

British thermal unit (ISO) | Show source$BTU_{ISO}$ | BTU_{ISO} | Show source$\text{...}$ | - | An obsolete heat unit defined by ISO 31-4 standard. The unit was created by rounding up the British thermal IT unit. See the BTU _{IT}unit for more information.$1\ BTU_{ISO} \equiv 1.05506\ kJ$ | Show source$...$ |

British thermal unit (63 °F) | Show source$BTU_{63^\circ F}$ | BTU_{63 °F} | Show source$\text{...}$ | - | A heat unit used in Anglo-Saxon countries. One British thermal unit defined for temperature sixty-three degrees fahrenheit (63°F) corresponds to the amount of energy needed to heat one pound of water (1 lb) from sixty-three (63°F) to sixty-four (64°F) degrees fahrenheit under pressure of one atmosphere (1 atm).$1\ BTU_{TH} = \Delta Q_{63 \rightarrow 64^{\circ}F} \approx 1.0546\ kJ$ | Show source$...$ |

British thermal unit (60 °F) | Show source$BTU_{60^\circ F}$ | BTU_{60 °F} | Show source$\text{...}$ | - | A heat unit used in the Canada. One British thermal unit defined for temperature sixty degrees fahrenheit (60°F) corresponds to the amount of energy needed to heat one pound of water (1 lb) from sixty (60°F) to sixty-one (61°F) degrees fahrenheit under pressure of one atmosphere (1 atm).$1\ BTU_{TH} = \Delta Q_{60 \rightarrow 61^{\circ}F} \approx 1054.68\ kJ$ | Show source$...$ |

British thermal unit (59 °F) | Show source$BTU_{59^\circ F}$ | BTU_{59 °F} | Show source$\text{...}$ | - | A heat unit used in United States for natural gas pricing. One British thermal unit defined for temperature fifty-nine degrees fahrenheit (59°F) corresponds to the amount of energy needed to heat one pound of water (1 lb) from fifty-nine (59°F) to sixty (60°F) degrees fahrenheit under pressure of one atmosphere (1 atm).$1\ BTU_{TH} = \Delta Q_{59 \rightarrow 60^{\circ}F} \approx 1054.80\ kJ$ | Show source$...$ |

British thermal unit (International Table) | Show source$BTU_{IT}$ | BTU_{IT} | Show source$\text{...}$ | - | A heat unit standardized during Fifth International Conference on Water and Steam Properties (IAPWS), which took place in 1956 in London. It was an attempt to organize various definitions of heat units by introducing the so-called an international British thermal unit exactly equal to 1.05505585262 kilojoules. See the other BTU unit definitions for more information.$1\ BTU_{IT} \equiv 1.05505585262\ kJ$ | Show source$...$ |

British thermal unit (mean) | Show source$BTU_{mean}$ | BTU_{mean} | Show source$\text{...}$ | - | Obsolete heat unit. One average British thermal unit corresponds to the average amount of energy needed to heat one pound of water (1 lb) by one degree of fahrenheit. Unit is defined as 1/180 of the heat needed to bring water from melting point (32°F) to boiling (212°F).$1\ BTU_{\text{śr.}} = \frac{1}{180} \Delta Q_{32 \rightarrow 212^{\circ}F} \approx 1.05587\ kJ$ | Show source$...$ |

British thermal unit (39 °F) | Show source$BTU_{39^\circ F}$ | BTU_{39} | Show source$\text{...}$ | - | A heat unit used in Anglo-Saxon countries. One British thermal unit defined for temperature thirty-nine degrees fahrenheit (39°F) corresponds to the amount of energy needed to heat one pound of water (1 lb) from thirty-nine (39°F) to fourty (40°F) degrees fahrenheit under pressure of one atmosphere (1 atm).$1\ BTU_{TH} = \Delta Q_{39 \rightarrow 40^{\circ}F} \approx 1.05967\ kJ$ | Show source$...$ |

Celsius heat unit (International Table) | Show source$CHU_{IT}$ | CHU_{IT} | Show source$\text{...}$ | - | An obsolete heat unit used in Anglo-Saxon countries. One Celsius heat unit (1 chu) corresponds to the amount of energy needed to heat one pound of water (1 lb) under pressure of one atmosphere (1 atm) by one degree Celsius (1°C). $1\ chu \approx 1.89918\ kJ$ | Show source$...$ |

quad | Show source$-$ | - | Show source$\text{...}$ | - | Unit energy used to describe world power industry. One quad corresponds to one quadrillion of british thermal units (10^{15} BTU). See the BTU unit for more.$1\ \text{quad} = 10^{15}\ BTU_{IT}$ | Show source$...$ |

therm (U.S.) | Show source$-$ | - | Show source$\text{...}$ | - | Energy unit used by natural gas companies in the United States. One americal therm (1 therm US) is defined as one hundred thousand of british thermal units for temperature of fifty-ninte deegres Fahrenheit (100,000 BTU_{59°F}), which approximately equals to amount of energy released while burning one hundred cubic feet of natural gas (100 cu ft). See the BTU unit for more.$1\ \text{therm (US)} = 100\ 000\ BTU_{\text{59°F}}$ | Show source$...$ |

therm (E.C.) | Show source$-$ | - | Show source$\text{...}$ | - | Therm unit used by engineers. One engineer therm (1 therm EC) equals to one one hundred thousand of international british thermal units (100,000 BTU_{IT}). See the therm (US) oraz BTU units for more.$1\ \text{therm (EC)} = 100\ 000\ BTU_{IT}$ | Show source$...$ |

# Calories related#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

calorie (20 °C) | Show source$cal_{20^\circ C}$ | cal_{20 °C} | Show source$\text{...}$ | - | Equivalent of energy needed to heat one gram (1 g) of water with temperature twenty degrees Celsius (20°C) under one atmosphere pressure (1 atm) by one degree (1°C).$1\ cal_{20^{\circ}C} = \Delta Q_{20 \rightarrow 21^{\circ}C} \approx 4.182\ J$ | Show source$...$ |

calorie (thermochemical) | Show source$cal_{th}$ | cal_{th} | Show source$\text{...}$ | - | An attempt to sort out various calorie definitions. One thermodynamic calorie corresponds exactly to 4.184 joules. See the other calorie definitions for more information.$1\ cal_{TH} \equiv 4.184\ J$ | Show source$...$ |

calorie (15 °C) | Show source$cal_{15^\circ C}$ | cal_{15 °C} | Show source$\text{...}$ | - | Equivalent of energy needed to heat one gram (1 g) of water with temperature fifteen degrees Celsius (15°C) under one atmosphere pressure (1 atm) by one degree (1°C).$1\ cal_{15^{\circ}C} = \Delta Q_{15 \rightarrow 16^{\circ}C} \approx 4.1855\ J$ | Show source$...$ |

calorie (International Table) | Show source$cal_{IT}$ | cal_{IT} | Show source$\text{...}$ | - | Heat unit standardized during Fifth International Conference on Water and Steam Properties (IAPWS), which took place in 1956 in London. Unit is an attempt to organize various calorie definitions by introducing so-called international calorie corresponding to exactly 4.868 of joule. See other calorie definitions to learn more.$1\ cal_{IT} \equiv 4.1868\ J$ | Show source$...$ |

calorie (mean) | Show source$cal_{mean}$ | cal_{mean} | Show source$\text{...}$ | - | Average energy needed to heat one gram (1 g) of water under pressure of one atmosphere (1 atm) by one degree Celsius (1°C). Defined as one hundredth (1/100) of heat needed to bring water from the melting point (0°C) to boiling (100°C).$1\ cal_{mean} = \frac{1}{100}\ \Delta Q_{0 \rightarrow 100^{\circ}C} \approx 4.190\ J$ | Show source$...$ |

calorie (3.98 °C) | Show source$cal_{3.98^\circ C}$ | cal_{3.98 °C} | Show source$\text{...}$ | - | Equivalent of energy needed to heat one gram (1 g) of water with temperature 3.98 degrees Celsius (3.98°C) under one atmosphere pressure (1 atm) by one degree (1°C).$1\ cal_{3.98^{\circ}C} = \Delta Q_{3.98 \rightarrow 4.98^{\circ}C} \approx 4.2045\ J$ | Show source$...$ |

kilocalorie | Show source$kcal$ | kcal | Show source$\text{...}$ | - | Equivalent to one thousand of international calories (1000 cal_{IT}). Sometimes called also large calorie. See the calorie unit for more.$1\ kcal = 1000\ cal_{IT}$ | Show source$...$ |

large calorie | Show source$Cal$ | Cal | Show source$\text{...}$ | - | Alternative name for kilocalorie (1 kcal). See kilocalorie unit for more.$1\ Cal = 1\ kcal = 1000\ cal_{IT}$ | Show source$...$ |

thermie | Show source$th$ | th | Show source$\text{...}$ | - | Equivalent to one thousand kilocalories (1000 kcal) or one million international calories (1,000,000 cal_{IT}). See the calorie or kilocalorie units for more.$1\ th = 1000\ kcal = 1\ 000\ 000\ cal_{IT}$ | Show source$...$ |

# Displacement related (UK/US)#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

foot-poundal | Show source$\text{ft pdl}$ | ft pdl | Show source$\text{...}$ | - | An obsolete heat unit used in Anglo-Saxon countries. Equivalent to work done by a force of one poundal (1 pdl) by shifting the point of force application by one foot (1 ft) in a direction parallel to the direction of the force.$\begin{aligned}1\ ft\ pdl &= 1\ ft \cdot 1\ pdl =\\&= 0.3048\ m \cdot 0.138254954\ N =\\&= 0.0421401099792\ m \cdot N =\\&= 0.0421401099792\ J\end{aligned}$ | Show source$...$ |

foot-pound force | Show source$\text{ft lbf}$ | ft lbf | Show source$\text{...}$ | - | An obsolete heat unit used in Anglo-Saxon countries. Equivalent to work done by a one pound-force (1 lbf) by shifting the point of force application by one foot (1 ft) in a direction parallel to the direction of the force.$\begin{aligned}1\ ft\ lbf &= 1\ ft \cdot 1\ lbf =\\&= 0.3048\ m \cdot 4.448221615\ N =\\&= 1.355817948252\ m \cdot N =\\&= 1.355817948252\ J\end{aligned}$ | Show source$...$ |

inch-pound force | Show source$\text{in lbf}$ | in lbf | Show source$\text{...}$ | - | An obsolete heat unit used in Anglo-Saxon countries. Equivalent to work done by a one pound-force (1 lbf) by shifting the point of force application by one inch (1 in) in a direction parallel to the direction of the force.$\begin{aligned}1\ in\ lbf &= 1\ in \cdot 1\ lbf =\\&= 0.0254\ m \cdot 0.112984829021\ N =\\&= 0.112984829021\ m \cdot N =\\&= 1.355817948252\ J\end{aligned}$ | Show source$...$ |

# Pressure related (UK/US)#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

cubic foot of atmosphere | Show source$ft^3 \times atm$ | cu ft atm; scf | Show source$\text{...}$ | - | An energy unit used in Anglo-Saxon countries. Equivalent to the work to be done to compress gas with volume of one cubic foot (1 cu ft) at pressure of one atmospheres (1 atm).$\begin{aligned}1\ cu\ ft\ atm &= 1\ ft^3 \cdot 101325\ Pa =\\&= 0.028316847\ m^{\cancel{3}} \cdot 101325 \frac{N}{\cancel{m^2}} =\\&= 2869.204522275\ m \cdot N =\\&= 2.869204522275\ kJ\end{aligned}$ | Show source$...$ |

cubic yard of atmosphere | Show source$yd^3 \times atm$ | cu yd atm; scy | Show source$\text{...}$ | - | An energy unit used in Anglo-Saxon countries. Equivalent to the work to be done to compress gas with volume of one cubic yard (1 cu yd) at pressure of one atmospheres (1 atm).$\begin{aligned}1\ cu\ yd\ atm &= 1\ ft^3 \cdot 101325\ Pa =\\&= 0.764554858\ m^{\cancel{3}} \cdot 101325 \frac{N}{\cancel{m^2}} =\\&= 78233.07584485\ m \cdot N =\\&= 78.23307584485\ kJ\end{aligned}$ | Show source$...$ |

gallon-atmosphere (US) | Show source$\text{US gal atm}$ | US gal atm | Show source$\text{...}$ | - | An energy unit used in Anglo-Saxon countries. Equivalent to the work to be done to compress gas with volume of one US gallon (1 gal US) at pressure of one atmospheres (1 atm).$\begin{aligned}1\ gal(US)\ atm &= 0.003785412\ m^3 \cdot 101325\ Pa =\\&= 383.5568709\ m \cdot N = 383.5568709\ J\end{aligned}$ | Show source$...$ |

gallon-atmosphere (imperial) | Show source$\text{imp gal atm}$ | imp gal atm | Show source$\text{...}$ | - | An energy unit used in Anglo-Saxon countries. Equivalent to the work to be done to compress gas with volume of one imperial gallon (1 gal UK) at pressure of one atmospheres (1 atm).$\begin{aligned}1\ gal(UK)\ atm &= 0.00454609\ m^3 \cdot 101325\ Pa =\\&= 460.63256925\ m \cdot N = 460.63256925\ J\end{aligned}$ | Show source$...$ |

# Pressure based (metric)#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

cubic centimetre of atmosphere | Show source$\text{cc atm; scc}$ | cc atm; scc | Show source$\text{...}$ | - | Equivalent to the work to be done to compress gas with volume of one cubic centimeter (1 cm³) at pressure of one atmospheres (1 atm). Sometimes called also standard cubic centimetre.$\begin{aligned}1\ cc\ atm &= 1\ cm^3 \cdot 101325\ Pa =\\&= 10^{-6}\ m^{\cancel{3}} \cdot 101325 \frac{N}{\cancel{m^2}} =\\&= 0.101325\ m \cdot N =\\&= 0.101325\ J\end{aligned}$ | Show source$...$ |

litre-atmosphere | Show source$\text{l atm}$ | l atm | Show source$\text{...}$ | - | Equivalent to the work to be done to compress gas with volume of one litre (1 l) at pressure of one atmospheres (1 atm).$\begin{aligned}1\ l\ atm &= 1\ dm^3 \cdot 101325\ Pa =\\&= 0.001\ m^{\cancel{3}} \cdot 101325 \frac{N}{\cancel{m^2}} =\\&= 101.325\ m \cdot N =\\&= 101.325\ J\end{aligned}$ | Show source$...$ |

# physical#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

atomic unit of energy | Show source$au$ | au | Show source$\text{...}$ | - | A unit of energy often used in quantum mechanical calculations. One atomic energy unit corresponds to two electron energies in a hydrogen atom in its ground state. Another name for this unit is hartree (1 Eh) or hartree's energy.$1\ au = 1\ E_h = \frac{e^2}{4 \pi \epsilon_0 a_0} = 4.359\ 743\ 81(34) \cdot 10^{-18}\ J$Where: - e - elementary charge,
- $\pi$ - math constant equals to about 3.14,
- $\epsilon_0$ - electric permeability of the vacuum,
- $a_0$ - radius of Bohr's first orbit.
| Show source$...$ |

hartree | Show source$E_h$ | E_{h} | Show source$\text{...}$ | - | Another name for atomic unit of energy. See atomic unit of energy for more.$1\ E_h = 2\ Ry$ | Show source$...$ |

electronvolt | Show source$eV$ | eV | Show source$\text{...}$ | - | A unit of energy used in various fields of physics and chemistry. One electronvolt (1 eV) corresponds to the energy that an electron receives or loses during acceleration within electric field with a potential difference of one volt (1 V). To calculate the value of one electronvolt in joules, we can multiply elementary charge (charge of single electron) by one volt.$\begin{aligned}1\ eV &= e \cdot 1\ V =\\&= 1.6021766208(98) \cdot 10^{-19}C \cdot 1\ \frac{W}{A} =\\&= 1.6021766208(98) \cdot 10^{-19}\ \cancel{A \cdot s} \cdot \frac{J}{\cancel{s \cdot A}} =\\&= 1.6021766208(98) \cdot 10^{-19}\ J\end{aligned}$ | Show source$...$ |

kilojoule per mol | Show source$\frac{kJ}{mol}$ | kJ/mol | Show source$\text{...}$ | - | Unit of energy per amount of substance unit. Widely used in thermodynamics to determine the energy of chemical reactions or phase transitions, e.g. enthalpy of evaporation.$1\ \frac{kJ}{mol}= \frac{1}{N_A}\ kJ = \frac{1000\ J}{6.02214076 \cdot 10^{23}} = 1.66053906717385 \cdot 10^{-21}\ J$Where: - $N_A$ - Avogadro constant equals to number of particles (atoms, molecules, ions etc.) in one mole of substance.
| Show source$...$ |

erg (cgs unit) | Show source$erg$ | erg | Show source$\text{...}$ | - | Historic energy unit in centimeter-gram-second system (CGS). One erg corresponds to the work done by force of one dyne (1 dyne) when the point of force application is shifted by one centimeter (1 cm) in a direction parallel to the direction of force.$1\ erg = 1\ dyn \cdot 1\ cm = \frac{10^{-3}\ kg \cdot 10^{-4}\ m}{s^2} = 10^{-7}\ J$ | Show source$...$ |

rydberg | Show source$Ry$ | Ry | Show source$\text{...}$ | - | A unit of energy used in atomic physics. One rydberg (1 Ry) corresponds to ionization energy of a hydrogen atom in the ground state.$1\ Ry = \frac{1}{2}\ E_h = \frac{e^2}{8 \pi \epsilon_0 a_0} = 2.179\ 871\ 905(17) \cdot 10^{-18}\ J$Where: - e - elementary charge,
- $\pi$ - math constant equals to about 3.14,
- $\epsilon_0$ - electric permeability of the vacuum,
- $a_0$ - radius of Bohr's first orbit.
| Show source$...$ |

# time related#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

horsepower-hour | Show source$hp \times h$ | hp·h | Show source$\text{...}$ | - | Amount of work done by an one horsepower engine (1 hp) within one hour (60 min).$\begin{aligned}1\ hp(I) \cdot h &= 745.69987158227022\ W \cdot 60\ min =\\&= 745.69987158227022\ \frac{J}{\cancel{s}} \cdot 3600\ \cancel{s} =\\&= 2.68451953769617\ MJ\end{aligned}$ | Show source$...$ |

watt-second | Show source$W \times s$ | W·s | Show source$\text{...}$ | - | Equivalent to one joule (1 J). One watt-second (1 W·s) corresponds to the amount of electric energy consumed (or approximately emited in the form of heat) by a device with power of one watt (1 W) within one second (1 s). See the joule unit to learn more.$1\ W \cdot s = 1\ \frac{J}{\cancel{s}} \cdot \cancel{s} = 1\ J$ | Show source$...$ |

kilowatt-second | Show source$kW \times s$ | kW·s | Show source$\text{...}$ | - | Equivalent to one kilojoule (1 kJ) or one thousand joules (1000 J). One kilowatt-second (1 kW·s) corresponds to the amount of electric energy consumed (or approximately emited in the form of heat) by a device with power of one kilowatt (1 kW) within one second (1 s). See the joule unit to learn more.$1\ kW \cdot s = 1\ \frac{1000\ J}{\cancel{s}} \cdot \cancel{s} = 1000\ J = 1\ kJ$ | Show source$...$ |

watt-hour | Show source$W \times h$ | W·h | Show source$\text{...}$ | - | Unit used to measure electricity consumption. One watt-hour (1 W·h) corresponds to the amount of electric energy consumed (or approximately emited in the form of heat) by a device with power of one watt (1 W) within one hour (60 min).$1\ W \cdot h = 1\ W \cdot 60\ min = 1\ \frac{J}{\cancel{s}} \cdot 3600\ \cancel{s} = 3.6\ kJ$ | Show source$...$ |

kilowatt-hour | Show source$kW \times h$ | kW·h | Show source$\text{...}$ | - | Unit used for electric cost pricing. One kilowatt-hour (1 kW·h) corresponds to the amount of electric energy consumed (or approximately emited in the form of heat) by a device with power of one kilowatt (1 kW) within one hour (60 min).$1\ kW \cdot h = 1000\ W \cdot 60\ min = 1000\ \frac{J}{\cancel{s}} \cdot 3600\ \cancel{s} = 3.6\ MJ$ | Show source$...$ |

# materials related#

Unit | Symbol | Symbol (plain text) | Value as symbolic | Value as numeric | Notes | Unit conversion formula |

barrel of oil equivalent | Show source$BOE$ | BOE | Show source$\text{...}$ | - | A unit of energy used in the power industry. Burning one barrel (42 US gallons) of crude oil (1 BOE) corresponds to the release of about six million British thermal units (5,800,000 BTU). See the BTU unit for more information.$1\ BOE = 5.8 \cdot 10^6\ BTU_{59^{\circ}F} = 6.1178632\ GJ$ | Show source$...$ |

ton of TNT | Show source$tTNT$ | tTNT | Show source$\text{...}$ | - | A unit used to determine amount of energy released in an explosion, e.g. to compare nuclear weapons. The explosion of one ton of TNT (1 tTNT) corresponds to release of about four gigajoules of energy (4 GJ). See the joule unit to learn more.$1\ tTNT = 4.184\ GJ$ | Show source$...$ |

ton of coal equivalent | Show source$TCE$ | TCE | Show source$\text{...}$ | - | A unit of energy used in the power industry. Burning one ton of coal (1 TCE) corresponds to release about twenty-nine gigajoules of energy (29 GJ). See the joule unit to learn more.$1\ TCE = 29.3076\ GJ$ | Show source$...$ |

ton of oil equivalent | Show source$TOE$ | TOE | Show source$\text{...}$ | - | A unit of energy used in the power industry. Burning one ton of crude oil (1 TOE) corresponds to release of about forty-two gigajoules of energy (42 GJ). See the joule unit to learn more.$1\ TOE = 41.868\ GJ$ | Show source$...$ |

cubic foot of natural gas | Show source$-$ | - | Show source$\text{...}$ | - | Equivalent of amount of energy released while burning out one cubic foot (1 cu ft) of natural gas.$1\ ft^3\ \text{natural gas} \approx 1000\ BTU_{59^{\circ}F}$ | Show source$...$ |

# Some facts#

- Energy is the
**scalar**physical quantity expressing the**ability to do the work**. - Energy is
**additive**. This means that the total energy of the system consisting of the N objects, is the sum of the energy of each of the bodies. - The kinetic energy is work to be done in order to provide the body with mass m, velocity V. It amounts to:

$E_{kin.} = \dfrac{m \times V^2}{2}$where:

- $E_{kin.}$ is the kinetic energy,

- $m$ is the mass,

- $V$ is the value of the velocity vector.

- $E_{kin.}$ is the kinetic energy,
- The potential energy at the point $\vec{x_0}$ is work to be done to put the body at this point (moving them from infinity).

- There are many different symbols used for potential energy depending on kind of science. Most common are U, V, or simply E
_{pot.}.

- Potential energy can be negative. This means that we don't need to perform the work to put the body in the current positions at all, but also it is needed to do the work to corrupt current system. In this case we say that
**system is in a bound**. A good example here are chemical molecules that are associated systems, because we need to do work to break chemical bonds.

- The function $U=f(\vec{x})$, which assigns value of potential energy to each point
**x**is commonly called**potential energy surface**. Sometimes, when people want to mark that surface have more than 3 dimensions (degree of freedom), they use term**hipersurface**. The concept of (hiper)surface of potential energy is widely used for example in**quantum chemistry**or**physics of the atomic nucleus**.

- There are many different symbols used for potential energy depending on kind of science. Most common are U, V, or simply E
- There are many forms of energy for example: heat or electrical.
- The basic energy unit in SI system is
**1J (one jul)**, so it's the same as unit of work. However, for practical reasons many different units are used depending on kind of science for example:

**elektronovolts (eV) in high-energy physics**,

**atomic units (au)**in quantum chemistry,

- calories in dietetic,

**horsepower**in automotive industry.

- The average kinetic energy of single particle divided by the number of degrees of freedom is
**temperature of the system**. Such concepts owe the development of statistical thermodynamics (physics), which made it possible to link the micro state (individual particles level) with macroscopic quantities (such as temperature, pressure). Previously, the concept of micro and macroscopic were independent. It is worth noting that the concept of temperature has**only statistical meaning**. This means for example that temperature for single particle has no meaning. - One of the fundamental laws of nature is the desire to minimize energy. There are no known causes of this fact, but an enormous amount of physical theory is based on this postulate. Very often the solution to a practical problem boils down to mininimalization energy problem. Examples include:

- Molecular mechanics - the way of finding optimal molecule geometry using clasical Newton dynamic.

- Variational methods - the set of general methods, that searches for wave functions, for which the system gives minimal average energy (formally the
**average value of the Hamiltonian**). Good examples are**Hartree-fock equations**, which (together with**Density Functional Theory - DFT**) are the foundations of modern quantum-mechanical calculations.

- Chemical reaction paths - sets of methods trying to search for optimal trace on energy surface.

From a mathematical point of view, that are classic optimization problems. Mathematical apparatus that deals with this kind of problem is - depending on whether we are looking for the numbers or functions -**calculus**or**calculus of variations**. - Molecular mechanics - the way of finding optimal molecule geometry using clasical Newton dynamic.
- If we have the potential energy surface, we can get forces that operate in various points in the system. To do this we need to calculate the energy derivative dE/dx in point. This fact is due to the reversal of the definition of work (integral of the product of the displacement and the applied force). Such a procedure may be used for numerical optimization of the geometry of the system. To do this we need to repeat in loop (as long as there are forces in the system):

- Compute forces working for each particle by computing derivate:

$\vec{F_0} = \dfrac{\partial{E}}{\partial{\vec{x_0}}}$ - Move particles by computed forces.

- Compute forces working for each particle by computing derivate:

# How to convert#

**Enter the number to field "value"**- enter the NUMBER only, no other words, symbols or unit names. You can use dot (**.**) or comma (**,**) to enter fractions.

Examples:- 1000000
- 123,23
- 999.99999

**Find and select your starting unit in field "unit"**. Some unit calculators have huge number of different units to select from - it's just how complicated our world is...**And... you got the result**in the table below. You'll find several results for many different units - we show you all results we know at once. Just find the one you're looking for.

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