# Heat and Resistance (1 Viewer)

#### Nailgun

##### Cole World
Just a question that came to me while I was writing notes
If resistance in a conductor causes electrical energy to turn into heat energy and when a conductor is heated it's resistance increases, wouldn't this be like an infinite loop if you tried to run electricity through a circuit in a vacuum, as in as the resistor turns the electricity into heat, it resists more electricity and hence more heat. Can the heat dissipate in space?

I assume the reason this doesn't happen in practical situations is because the heat is dissipated into the air?

#### leehuan

##### Well-Known Member
Power loss = R I^2 -------> Energy loss = R I^2 t

Actually, if you have more CURRENT running through, and the material DOES have resistance, then it affects heat losses more.

(Aside: However, if you have a perfect conductor that has 0 resistance you will never experience heat losses)
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As for the main question, I think it's a simple matter of fact that according to conservation of mass-energy, mass can also be transFERRED as well as transmitted. It's a form of energy; you don't exactly need particles present in a vacuum to transfer energy around. The one you need particles to transfer is a sound wave.

#### Nailgun

##### Cole World
But don't you need a medium to transmit energy unless its electromagnetic radiation?

I think you might of misunderstood my question (or I misunderstood your answer)
What I mean is that if you have a circuit with a resistance R, as the current passes through the circuit energy is lost to heat as shown in your formula yeah?
I've read that resistance of a substance increases with heat as the atoms are more likely to collide with electrons if they're oscillating everywhere, so as energy is lost from the current in the form of heat, wouldn't this heat then heat up the circuit which would increase the resistance of the circuit and therefore further energy loss.

So I think it would be like a self-perpetuating cycle where the energy is lost at a increasing rate as the temperature of the resistor increases (and hence the resistance). Basically my question does this happen, and if not, why?

#### leehuan

##### Well-Known Member
"Radiation is the heat that we feel coming from a hot object. It warms the air using heat waves (infrared waves) that radiate out from the hot object in all directions until it is absorbed by other objects. Transfer of heat by radiation travels at the speed of light and goes great distances."

Source: http://sciencelearn.org.nz/Contexts/Fire/Science-Ideas-and-Concepts/Heat-energy

It's obviously not convection. But if we're in a vacuum minus the energy source, what's there to conduct?

#### leehuan

##### Well-Known Member
Scenario: Electrons are travelling through the lattice. Resistance is when the electrons essentially hit some kind of obstacle (this could be the lattice itself, or say an impurity).

It's obviously not fast enough to fuse an electron and a proton into a neutron. But the electron just gets bounced off, becomes a bit slower, and the heat energy loss just gets radiated away

#### Nailgun

##### Cole World
Right, but wouldn't at least some of that heat be absorbed by the conductor, and hence increase the resistance of the conductor?

#### leehuan

##### Well-Known Member
You raise a valid point which I'll let someone else address but I thought P=I^2R caters for only the radiated away cause anything absorbed would just get reconverted?

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#### anomalousdecay

You need a very high current in order to increase the heat a lot. It depends on the wires used as well. In most practical situations, you can avoid this by using thick wires.

The thicker the wiring and depending on the material and type of wire, the lower the resistance of the conductor.

In industry, different wires will have a specification for maximum current. Once you reach a certain specified current, you can indeed tend to heat up the wire. There are safe ratings given but there is also a point called the "fusing current" where the heat goes up too high and the wires can burn/melt.

Again, depending on the wire material and size and the location (ie it could be in open wind or it could be in a machine at over 500 degrees celsius), the resistance will increase at a different rate due to heat. However, as the resistance increases, if the voltage is fixed (for example, a 5 V fixed voltage source), then the current will decrease. The power lost due to heat is given by $\bg_white P = I^2 R$.

Since the power lost is proportional to the square of the current, then the larger the current the more power losses occur compared to an increase in current.

As long as the wire is thick enough and made from a good material for the application, it is fine to withstand larger currents (and hence doesn't experience as high power losses) as the resistance stays sufficiently low enough. If it heats up and the resistance increases, the maximum current decreases so the power losses aren't as high anymore if the voltage is fixed (which it is in most cases).

#### Nailgun

##### Cole World
You raise a valid point which I'll let someone else address but I thought P=I^2R caters for only the radiated away cause anything absorbed would just get reconverted?

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I have no idea lol, I guess you'd have to do that derivation to tell.
Even then though, regardless of the formula the heat would still hypothetically increase - the formula just tells you how much is lol
Would a circuit stabilize somehow eventually, or does the resistance just increase as long as energy is being pumped in from a power source.

That being said, I thought that electromagnetic radiation is only generated when a charge is accelerated, how does the heat radiate then? Does it accelerate at some point

#### anomalousdecay

You raise a valid point which I'll let someone else address but I thought P=I^2R caters for only the radiated away cause anything absorbed would just get reconverted?
It is dependent on the environment. Heat energy can be transferred over but it still depends on the environment. The whole system will attempt to reach thermal equilibrium.

$\bg_white P = I^2 R$ just tells you how much electrical energy is transformed into heat energy.

#### Nailgun

##### Cole World
Oh this makes a lot of sense.
Basically what you're saying is that the resistance is inversely proportional to the current because of Ohms Law
Then even if the resistance increases due to heat, the power loss will be reduced because of a lower current

Okay, I think I need to try do this mathematically to really get a grip on like the relationship. Is that like possible lol, and how would you recommend I start?

#### InteGrand

##### Well-Known Member
Oh this makes a lot of sense.
Basically what you're saying is that the resistance is inversely proportional to the current because of Ohms Law
Then even if the resistance increases due to heat, the power loss will be reduced because of a lower current

Okay, I think I need to try do this mathematically to really get a grip on like the relationship. Is that like possible lol, and how would you recommend I start?
Yeah, in fact, nothing in (real) physics can be understood properly without mathematics.

Btw, resistance generally increases linearly with temperature, see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/restmp.html

#### anomalousdecay

Make sure you understand the difference between a voltage source and voltage drop.

The reason why you get the power loss across a resistance as $\bg_white P = I^2 R$ is because the voltage drop across the resistance is V = IR. Electrical power is always given by P = VI. But, because there is a voltage drop the power loss is given by $\bg_white P_{loss} = V_{drop} I = I^2 R$

#### anomalousdecay

Btw, resistance generally increases linearly with temperature, see here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/restmp.html
Not necessarily. There might be a region where the relationship between resistance and temperature is linear however it depends on many different factors, including the material and hence the properties of that specific wire (thickness, current ratings, etc).

Also when you are closer to the fusing current (this happens when there is a fixed current load/source), the wire heats up very quickly and as a result it burns up and open circuits.

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#### InteGrand

##### Well-Known Member
Not necessarily. There might be a region where the relationship between resistance and temperature is linear however it depends on many different factors, including the material and hence the properties of that specific wire (thickness, current ratings, etc).

Also when you are closer to the fusing current (this happens when there is a fixed current load/source), the wire heats up very quickly and as a result it burns up and open circuits.
I'm guessing the resistance is linear vs. temperature when the temperature is near the reference temperature in that formula? But far away from this, it could be a steeper curve?

#### Nailgun

##### Cole World
Okay so I tried so see if I could work out a relationship
$\bg_white Take\quad a\quad circuit\quad with\quad a\quad current\quad I\quad and\quad supply\quad voltage\quad V\quad with\quad a\quad resistor\quad with\quad resistance\quad r\quad \Omega \\ Over\quad resistor\quad V_{ DROP }=I.r\\ Hence\quad P_{ DROP }={ I }^{ 2 }r\\ P_{ DROP }.z\quad =\quad Q\quad (Let\quad z\quad be\quad coefficent\quad of\quad how\quad much\quad heat\quad energy\quad is\quad absorbed)\\ Q=mC\Delta T\quad (Is\quad this\quad formula\quad appropriate?)\\ Hence,\quad \Delta T=\frac { Q }{ mC } \quad (m\quad is\quad mass\quad of\quad conductor?\quad C\quad is\quad specific\quad heat\quad capactiy\quad of\quad conductor)\\ \frac { \Delta R }{ r } =\alpha \Delta T\\ \frac { \Delta R }{ r } =\frac { { I }^{ 2 }rz }{ mC } \\ \Delta R=\frac { { I }^{ 2 }{ r }^{ 2 }z }{ mC }$
But not sure where to go with this, I assume calculus is needed given this is like rate of change but for what?
I'm thinking dP/dt? I know that Power is measured in J/sec though, so wouldn't it just be like linear

#### Nailgun

##### Cole World
Wait what I'm looking for is a relationship for what P lost is equal to after n amount of time
Will P lost > P eventually I think?

#### anomalousdecay

I'm guessing the resistance is linear vs. temperature when the temperature is near the reference temperature in that formula? But far away from this, it could be a steeper curve?
One reason is that the circuit could have several different components in it with different temperature coefficients of resistance.

However, the coefficient itself is not a fixed constant for the material. So if you still only consider the one element, you can only approximate the relationship. It is dependent on the temperature that the material is currently at.

#### anomalousdecay

You can use $\bg_white Q = mC \Delta T$ ONLY for the case when heat is transferred over convection or conduction. However you also have to take into account the thermal equilibrium of the whole system, including the environment (if there is no wind, then you can take final temperature as ambient air temperature of the room).

You can look at this two ways. I will explain both below.

For the case where the conductor is cooler than the ambient temperature: The value you get for Q is the amount of heat energy required to heat up the conductor to the ambient temperature (if necessary). m is the mass of the conductor, C is specific heat of the conductor and delta T is the change in temperature required for the conductor and room to reach thermal equilibrium. As long as the current isn't too high as so that the conductor is beginning to burn, the conductor will stay at this ambient temperature as the room will act as a heat dissipator.

For the case where the conductor is at a higher temperature than the ambient temperature: The value you get for Q is the amount of heat energy required to heat up the room to the temperature of the conductor. m is the mass of the gas/liquid/solid (remember, this is only for convection or conduction heat transfer!) within the proximity of the wire (ie the environment), C is the specific heat of the gas/liquid/solid and delta T is the change in temperature required for the environment and conductor to reach thermal equilibrium. In this case, the room will increase in temperature until the equilibrium is established. Effectively, for this whole process the room will again act as a heat dissipator.

For both of these conditions, the following will occur after equilibrium.

There is more electrical energy converted into heat energy in this system after this stage. So you ask, what is happening to this energy?

The whole system will rise in temperature. The differences in temperature will depend on the heat transfer rate from the wire to the environment which is dependent on many more various factors.

Think of this case as the same thing as a CPU and CPU heat sink. Imagine the wire as the CPU, whilst the environment is analogous to the heat sink. The heat sink will rise in temperature until it reaches the same temperature as the CPU. Then, they will both steadily increase in temperature. What saves this system from burning you ask? The fan for dissipating heat into the air. Generally, a good fan will ensure that the heat sink is cooled very quickly. If not, when the CPU returns to a low load and is actually generating less heat, the heat sink will transfer heat back to the CPU meaning the fan on the heat sink is not doing it's proposed job and is very bad for the application. You can also apply this analogy to a radiator core which can be found on a CPU or car engine as well for liquid cooling.

You had the right idea for convection and conduction cases, but you didn't apply the theory properly.

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#### Nailgun

##### Cole World
You can use $\bg_white Q = mC \Delta T$ ONLY for the case when heat is transferred over convection or conduction. However you also have to take into account the thermal equilibrium of the whole system, including the environment (if there is no wind, then you can take final temperature as ambient air temperature of the room).

You can look at this two ways. I will explain both below.

For the case where the conductor is cooler than the ambient temperature: The value you get for Q is the amount of heat energy required to heat up the conductor to the ambient temperature (if necessary). m is the mass of the conductor, C is specific heat of the conductor and delta T is the change in temperature required for the conductor and room to reach thermal equilibrium. As long as the current isn't too high as so that the conductor is beginning to burn, the conductor will stay at this ambient temperature as the room will act as a heat dissipator.

For the case where the conductor is at a higher temperature than the ambient temperature: The value you get for Q is the amount of heat energy required to heat up the room to the temperature of the conductor. m is the mass of the gas/liquid/solid (remember, this is only for convection or conduction heat transfer!) within the proximity of the wire (ie the environment), C is the specific heat of the gas/liquid/solid and delta T is the change in temperature required for the environment and conductor to reach thermal equilibrium. In this case, the room will increase in temperature until the equilibrium is established. Effectively, for this whole process the room will again act as a heat dissipator.

For both of these conditions, the following will occur after equilibrium.

There is more electrical energy converted into heat energy in this system after this stage. So you ask, what is happening to this energy?

The whole system will rise in temperature. The differences in temperature will depend on the heat transfer rate from the wire to the environment which is dependent on many more various factors.

Think of this case as the same thing as a CPU and CPU heat sink. Imagine the wire as the CPU, whilst the environment is analogous to the heat sink. The heat sink will rise in temperature until it reaches the same temperature as the CPU. Then, they will both steadily increase in temperature. What saves this system from burning you ask? The fan for dissipating heat into the air. Generally, a good fan will ensure that the heat sink is cooled very quickly. If not, when the CPU returns to a low load and is actually generating less heat, the heat sink will transfer heat back to the CPU meaning the fan on the heat sink is not doing it's proposed job and is very bad for the application. You can also apply this analogy to a radiator core which can be found on a CPU or car engine as well for liquid cooling.

You had the right idea for convection and conduction cases, but you didn't apply the theory properly.
Okay that makes a lot of sense. So what happens is that the temperatures will become homogeneous in that area <-- is there a term for this tendency?
When equilibrium is reached, from that point the space as whole would heat up. So I guess material limitations aside, the loop would occur, but be extremely limited by the amount of current necessary to generate enough heat, and that much of the heat would be dissipated into the air which is acting like a heat-sink. That and that as resistance increases the current would decrease meaning eventually the process would slow down into almost negligible change.

Alright, I think I've got it mostly
Thanks All
Much appreciated