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Conservation of Mass and Energy

Conversion of mass to energy in chemical/nuclear reactionsCan non-free forces change the rest mass?Why does mass change in to energy during a nuclear change?Energy & Mass of a PhotonWhat is the argument for detailed balance in chemistry?Would impact angle matter on relativistic impactor?Hypothetical special relativity with mass conservationDoes the mass of object really increase?Some calculations on the energy consumption of a relativistic rocketIf mass and energy are same what will be the equivalent of a homogeneous ball in terms of energy and information?

1

$begingroup$

I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!

For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$

However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?

special-relativity conservation-laws mass-energy physical-chemistry

share|cite|improve this question

edited 56 mins ago

Aaron Stevens

12.8k42248

asked 1 hour ago

Dude156Dude156

1307

$endgroup$

add a comment | 

1

$begingroup$

I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!

For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$

However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?

special-relativity conservation-laws mass-energy physical-chemistry

share|cite|improve this question

edited 56 mins ago

Aaron Stevens

12.8k42248

asked 1 hour ago

Dude156Dude156

1307

$endgroup$

add a comment | 

$begingroup$

I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!

For example, consider combustion:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + {Energy}$$

However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?

special-relativity conservation-laws mass-energy physical-chemistry

share|cite|improve this question

edited 56 mins ago

Aaron Stevens

12.8k42248

asked 1 hour ago

Dude156Dude156

1307

$endgroup$

share|cite|improve this question

edited 56 mins ago

Aaron Stevens

12.8k42248

asked 1 hour ago

Dude156Dude156

1307

share|cite|improve this question

edited 56 mins ago

Aaron Stevens

12.8k42248

asked 1 hour ago

Dude156Dude156

1307

edited 56 mins ago

Aaron Stevens

12.8k42248

edited 56 mins ago

Aaron Stevens

12.8k42248

asked 1 hour ago

Dude156Dude156

1307

asked 1 hour ago

Dude156Dude156

1307

3 Answers
3

active

oldest

votes

2

$begingroup$

Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.

So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.

share|cite|improve this answer

edited 49 mins ago

answered 57 mins ago

F16FalconF16Falcon

3007

$endgroup$

add a comment | 

1

$begingroup$

It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.

In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.

share|cite|improve this answer

edited 13 mins ago

answered 1 hour ago

TechDroidTechDroid

60912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  scientificamerican.com/article/…
  $endgroup$
  – safesphere
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, Einstein has taken it all. But thanks for the link, it added something.
  $endgroup$
  – TechDroid
  12 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.

In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.

share|cite|improve this answer

answered 18 mins ago

PhysicsDavePhysicsDave

94547

$endgroup$

add a comment | 

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2

$begingroup$

Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.

So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.

share|cite|improve this answer

edited 49 mins ago

answered 57 mins ago

F16FalconF16Falcon

3007

$endgroup$

add a comment | 

2

$begingroup$

Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.

So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.

share|cite|improve this answer

edited 49 mins ago

answered 57 mins ago

F16FalconF16Falcon

3007

$endgroup$

add a comment | 

$begingroup$

Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.

So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.

share|cite|improve this answer

edited 49 mins ago

answered 57 mins ago

F16FalconF16Falcon

3007

$endgroup$

share|cite|improve this answer

edited 49 mins ago

answered 57 mins ago

F16FalconF16Falcon

3007

answered 57 mins ago

F16FalconF16Falcon

3007

answered 57 mins ago

F16FalconF16Falcon

3007

1

$begingroup$

It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.

In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.

share|cite|improve this answer

edited 13 mins ago

answered 1 hour ago

TechDroidTechDroid

60912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  scientificamerican.com/article/…
  $endgroup$
  – safesphere
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, Einstein has taken it all. But thanks for the link, it added something.
  $endgroup$
  – TechDroid
  12 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.

In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.

share|cite|improve this answer

edited 13 mins ago

answered 1 hour ago

TechDroidTechDroid

60912

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  scientificamerican.com/article/…
  $endgroup$
  – safesphere
  50 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, Einstein has taken it all. But thanks for the link, it added something.
  $endgroup$
  – TechDroid
  12 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.

In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.

share|cite|improve this answer

edited 13 mins ago

answered 1 hour ago

TechDroidTechDroid

60912

$endgroup$

share|cite|improve this answer

edited 13 mins ago

answered 1 hour ago

TechDroidTechDroid

60912

answered 1 hour ago

TechDroidTechDroid

60912

answered 1 hour ago

TechDroidTechDroid

60912

0

$begingroup$

All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.

In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.

share|cite|improve this answer

answered 18 mins ago

PhysicsDavePhysicsDave

94547

$endgroup$

add a comment | 

0

$begingroup$

All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.

In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.

share|cite|improve this answer

answered 18 mins ago

PhysicsDavePhysicsDave

94547

$endgroup$

add a comment | 

$begingroup$

All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.

In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.

share|cite|improve this answer

answered 18 mins ago

PhysicsDavePhysicsDave

94547

$endgroup$

share|cite|improve this answer

answered 18 mins ago

PhysicsDavePhysicsDave

94547

answered 18 mins ago

PhysicsDavePhysicsDave

94547

answered 18 mins ago

PhysicsDavePhysicsDave

94547