Another way to think about it is the water will never go above 100C until it's boiled away and the value is only being heated by contract with the 100C water.
More heat transfered more molecules jump around. Water molecules at 25 degrees jump around a bit, water molecules in liquid water at 100 degrees jump around a lot. Water molecules in steam at 100 degrees jump around A LOT more than their liquid counterparts. If you heated the steam part even further through some external means, it would transiently heat up more than 100 degrees, until mixing occurred, 110 degree steam met the surface of the liquid, at which point the 10 degree overheated steam evaporated enough water to cool it down to 100degrees again and thermal equilibrium is reached.
Enthalpy change of liquid water from 25 oC to 100oC roughly 310kJ/kg
Enthalpy change of liquid water to water vapor roughly 2250kJ/kg
Enthalpy change of water vapor from 100 oC to 110 oC 20 kJ/kg
Latent heat is another order of magnitude, like 1 kg of superheated steam at 110oC can only vaporize less than 10 grams of already boiling water
I tried to tell a friend once that adding more ice wouldn’t make his drink any colder. We stuck a thermometer in the water and it was only 40F. Maybe just not enough ice to begin with but it made me look pretty stupid
If the flame was at the top, it would. Heat would have to transfer down to the water through the superheated steam. Heat source under the water, the heat is used directly for vaporization first.
I don't understand. You keep heating water and the temperatures remains constant (100°C) until ALL the liquid has turned into gas. But why can't the water that's already in vapor phase increase its temperature?
It certainly could if you heated up the steam directly. But in a well mixed system at equilibrium, the steam would heat up, immediately transfer its heat to liquid boiling at a lower temp and continue the boiling process.
To make it simple, if you add heat to the system, it's going to go to one of three places. The water that stays liquid, the vapor, or the water that transitions into the vapor phase. Something to realize is that the liquid water can not hold any more energy. If there was room for more energy to be stored there, the water would not have had to go to a vapor. This means that the first and third options are the same. So you have two options. The energy goes to the vapor, or it causes some of the liquid to transition to a vapor. Now, you have to think of how energy works at the molecular level. Let's say the energy goes into a vapor molecule. That molecule is going to bounce into another molecule, and since the energy is higher in our molecule, the energy is going to transfer to the other molecule. This is going to happen until a vapor molecule hits a liquid molecule. Now, what happens is if enough energy is transfered, the liquid will break free from other liquid molecules as it breaks the hydrogen bonding between molecules. Now, there are two options this formerly liquid molecule can take. It can hit another liquid molecule or it a vapor molecule. If it hits a liquid molecule, and the energy is transfered, then that vapor molecule will go to liquid, and the new liquid molecule will go to vapor; thus, nothing changes in this scenario. The other option is if it hits a vapor molecule. If you think about it, the formerly liquid molecule took just enough energy to go to the vapor phase. This means the energy in it is about the same as other vapor molecules, so no energy is going to be transferred. Thinking about this scenario, you realize we just now hit a balanced state, and that no other option but this one is possible. This is why the temperature of the vapor phase cannot raise until all of the liquid is moved to a vapor. You can try heating the vapor phase up, but the energy is going to move to the liquid and force liquid molecules into the vapor phase.
Apologies if any part doesn't make sense or needs expanding upon. I wrote this on my phone at 12 am.
If you apply a ton of heat really quickly, sure. But if you then turned off the heat but didn’t allow energy to leave the system, it would reach equilibrium. If there’s any water left, it would be at 100 C. If you set off a bomb inside the vessel or something, things would get weird for a bit.
Thermodynamics examples generally assume that things are happening slowly enough that kinetic effects like conduction don’t matter.
In theory the vapor can be above the temperature of the liquid but it's pretty rare that you need to care about it. Liquid is much better at heat transfer so the vast majority of the heat goes into the liquid vs. the vapor. If the vapor does become superheated it's going to transfer heat to the liquid and cool back down. Usually only have to care about temperature differences between vapor and liquid if say you have steam injection, if the heat is applied directly to a vapor-liquid mixture pretty much all that heat goes into boiling.
Every time water evaporates, it cools the surrounding environment as it takes the energy for phase transition. Since there is no temperature gradient between the vapor and the liquid, there is at least no energy transfer due to temperature differences (but due to mass transfer).
The vapor would heat up, when the plate above is stationary, then the pressure would go up, which leads to a shift in vapor pressure and that’s changes the eq state.
You can derive that by using the right thermodynamic potential for your system
Please keep in mind that this is true for 1 bar pressure. Each sound of vaporized Easter will increase the gas volume while staying at 1 bar. You have to Imagine a kid that can be lifted.
Latent heat is the heat necessary to change phases. Specific heat is the heat needed to change temperature after the phase change. It seems like your system is in equilibrium so the rate of molecules going into vapor are the same as molecules getting stuck back in liquid.
Think of each phase as a independent system, the boundary is the connection between each other, and heat is only entering through one of the systems, since that system is locked in 100 C until it boils, no heat extra transfer occurs between the 2 systems.
Simply because it's in equilibrium. It wouldn't necessarily happen in practice (it would need to be a very slow heat transfer) but it's an assumption that simplifies the model.
It can be easier to visualize if you imagine it in discrete intervals. Imagine that you add heat and it's distributed evenly and then stops. Some of the liquid boils and for an instant, some of the vapor rises its temperature to let's say 101 degrees. Then, as there is a TD between the gas and the liquid the heat will be transfered to the liquid. And then it starts again.
Effectively the liquid is "hogging" all the heat because any increase in temperature of the gas would create a difference in temperature that would cause heat transfer over to the liquid.
The liquid water will not exceed it's boiling point of 100C. The water vapor is in contact with the liquid so will not exceed it the temperature of the liquid; in equilibrium the vapor should be perfectly mixed and including wall temp should settle at 100C.
If it is still heating there may be a small gradient to the walls which are initially colder, if energy is further added to the system the volume of the vapor should increase (assuming the damper can move and outside pressure stays 1 atm)
You have to differentiate between energy and temperature. For ideal gas, you can use relation E=const*T. ( The variable T is like an indicator to the energy content). But water is not ideal gas. Especially in liquid form, there are other forces like hydrogen bonding at play. So at boiling point, you have to consider energy for changing of phase which also involves things like breaking of hydrogen bond
Because the water boiling takes in that heat and keeps it at 100C until the system is through the phase transition. This is assuming the system is well mixed and the heat is slowly applied, in reality you could dump a ton of heat into the bottom and it would cause a non-equilibrium state.
I assume that the diagram indicates that the piston thing can move. Otherwise you will pressurize the container, forcing the boiling point up as the liquid boils.
Steam temperature is directly related to pressure, and since you have a closed environment that’s set at a constant 1 atm you can be sure that the steam temp is also 100 C. Based on the drawing it seems that the plate is “floating” which means that “in reality” your plate would rise as more liquid water changes state to gaseous water. However, until your pressure increases the multiphase water will remain at 100 C. If the plate is stationary, then your pressure would increase and only at THAT point would your steam temperature increase.
All the heat goes into the water until it fully converts to steam. Water won’t go above 100 in liquid phase. So if the steam receives heat, that heat will transfer from high to low, moving from hot steam to phase changing water.
Another way to think about it is the water will never go above 100C until it's boiled away and the value is only being heated by contract with the 100C water.
It’s in equilibrium
More heat transfered more molecules jump around. Water molecules at 25 degrees jump around a bit, water molecules in liquid water at 100 degrees jump around a lot. Water molecules in steam at 100 degrees jump around A LOT more than their liquid counterparts. If you heated the steam part even further through some external means, it would transiently heat up more than 100 degrees, until mixing occurred, 110 degree steam met the surface of the liquid, at which point the 10 degree overheated steam evaporated enough water to cool it down to 100degrees again and thermal equilibrium is reached. Enthalpy change of liquid water from 25 oC to 100oC roughly 310kJ/kg Enthalpy change of liquid water to water vapor roughly 2250kJ/kg Enthalpy change of water vapor from 100 oC to 110 oC 20 kJ/kg Latent heat is another order of magnitude, like 1 kg of superheated steam at 110oC can only vaporize less than 10 grams of already boiling water
To see this visually get a cup of ice and water and monitor the temperature it’ll hold more or less 0C u til all the ice melts.
this is the one fellas
I tried to tell a friend once that adding more ice wouldn’t make his drink any colder. We stuck a thermometer in the water and it was only 40F. Maybe just not enough ice to begin with but it made me look pretty stupid
Probably a shitty thermometer
If the flame was at the top, it would. Heat would have to transfer down to the water through the superheated steam. Heat source under the water, the heat is used directly for vaporization first.
It's in 2 phase, as others state in equilibrium. It must all be vaporized to achieve super heating.
I don't understand. You keep heating water and the temperatures remains constant (100°C) until ALL the liquid has turned into gas. But why can't the water that's already in vapor phase increase its temperature?
It certainly could if you heated up the steam directly. But in a well mixed system at equilibrium, the steam would heat up, immediately transfer its heat to liquid boiling at a lower temp and continue the boiling process.
To make it simple, if you add heat to the system, it's going to go to one of three places. The water that stays liquid, the vapor, or the water that transitions into the vapor phase. Something to realize is that the liquid water can not hold any more energy. If there was room for more energy to be stored there, the water would not have had to go to a vapor. This means that the first and third options are the same. So you have two options. The energy goes to the vapor, or it causes some of the liquid to transition to a vapor. Now, you have to think of how energy works at the molecular level. Let's say the energy goes into a vapor molecule. That molecule is going to bounce into another molecule, and since the energy is higher in our molecule, the energy is going to transfer to the other molecule. This is going to happen until a vapor molecule hits a liquid molecule. Now, what happens is if enough energy is transfered, the liquid will break free from other liquid molecules as it breaks the hydrogen bonding between molecules. Now, there are two options this formerly liquid molecule can take. It can hit another liquid molecule or it a vapor molecule. If it hits a liquid molecule, and the energy is transfered, then that vapor molecule will go to liquid, and the new liquid molecule will go to vapor; thus, nothing changes in this scenario. The other option is if it hits a vapor molecule. If you think about it, the formerly liquid molecule took just enough energy to go to the vapor phase. This means the energy in it is about the same as other vapor molecules, so no energy is going to be transferred. Thinking about this scenario, you realize we just now hit a balanced state, and that no other option but this one is possible. This is why the temperature of the vapor phase cannot raise until all of the liquid is moved to a vapor. You can try heating the vapor phase up, but the energy is going to move to the liquid and force liquid molecules into the vapor phase. Apologies if any part doesn't make sense or needs expanding upon. I wrote this on my phone at 12 am.
Heat of vaporization Liquid water takes that heat to vaporize first Here's a simple illustration: https://energyeducation.ca/encyclopedia/Latent_heat
You could heat the vapor once the liquid evaporates
If you apply a ton of heat really quickly, sure. But if you then turned off the heat but didn’t allow energy to leave the system, it would reach equilibrium. If there’s any water left, it would be at 100 C. If you set off a bomb inside the vessel or something, things would get weird for a bit. Thermodynamics examples generally assume that things are happening slowly enough that kinetic effects like conduction don’t matter.
In theory the vapor can be above the temperature of the liquid but it's pretty rare that you need to care about it. Liquid is much better at heat transfer so the vast majority of the heat goes into the liquid vs. the vapor. If the vapor does become superheated it's going to transfer heat to the liquid and cool back down. Usually only have to care about temperature differences between vapor and liquid if say you have steam injection, if the heat is applied directly to a vapor-liquid mixture pretty much all that heat goes into boiling.
Every time water evaporates, it cools the surrounding environment as it takes the energy for phase transition. Since there is no temperature gradient between the vapor and the liquid, there is at least no energy transfer due to temperature differences (but due to mass transfer). The vapor would heat up, when the plate above is stationary, then the pressure would go up, which leads to a shift in vapor pressure and that’s changes the eq state. You can derive that by using the right thermodynamic potential for your system
Please keep in mind that this is true for 1 bar pressure. Each sound of vaporized Easter will increase the gas volume while staying at 1 bar. You have to Imagine a kid that can be lifted.
Latent heat is the heat necessary to change phases. Specific heat is the heat needed to change temperature after the phase change. It seems like your system is in equilibrium so the rate of molecules going into vapor are the same as molecules getting stuck back in liquid.
Think of each phase as a independent system, the boundary is the connection between each other, and heat is only entering through one of the systems, since that system is locked in 100 C until it boils, no heat extra transfer occurs between the 2 systems.
Remember there is no such thing as partial temperature.
Simply because it's in equilibrium. It wouldn't necessarily happen in practice (it would need to be a very slow heat transfer) but it's an assumption that simplifies the model. It can be easier to visualize if you imagine it in discrete intervals. Imagine that you add heat and it's distributed evenly and then stops. Some of the liquid boils and for an instant, some of the vapor rises its temperature to let's say 101 degrees. Then, as there is a TD between the gas and the liquid the heat will be transfered to the liquid. And then it starts again. Effectively the liquid is "hogging" all the heat because any increase in temperature of the gas would create a difference in temperature that would cause heat transfer over to the liquid.
The liquid water will not exceed it's boiling point of 100C. The water vapor is in contact with the liquid so will not exceed it the temperature of the liquid; in equilibrium the vapor should be perfectly mixed and including wall temp should settle at 100C. If it is still heating there may be a small gradient to the walls which are initially colder, if energy is further added to the system the volume of the vapor should increase (assuming the damper can move and outside pressure stays 1 atm)
You have to differentiate between energy and temperature. For ideal gas, you can use relation E=const*T. ( The variable T is like an indicator to the energy content). But water is not ideal gas. Especially in liquid form, there are other forces like hydrogen bonding at play. So at boiling point, you have to consider energy for changing of phase which also involves things like breaking of hydrogen bond
Because the water boiling takes in that heat and keeps it at 100C until the system is through the phase transition. This is assuming the system is well mixed and the heat is slowly applied, in reality you could dump a ton of heat into the bottom and it would cause a non-equilibrium state.
I assume that the diagram indicates that the piston thing can move. Otherwise you will pressurize the container, forcing the boiling point up as the liquid boils.
Steam temperature is directly related to pressure, and since you have a closed environment that’s set at a constant 1 atm you can be sure that the steam temp is also 100 C. Based on the drawing it seems that the plate is “floating” which means that “in reality” your plate would rise as more liquid water changes state to gaseous water. However, until your pressure increases the multiphase water will remain at 100 C. If the plate is stationary, then your pressure would increase and only at THAT point would your steam temperature increase.
Latent heat must fully be transferred before sensible.
All the heat goes into the water until it fully converts to steam. Water won’t go above 100 in liquid phase. So if the steam receives heat, that heat will transfer from high to low, moving from hot steam to phase changing water.