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Physics Latent Heat

Can someone explain an experiment to determine the Specific Latent Heat of Vaporisation? Let's say for water.
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Original post by AlishaWhite
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You set an ice in a funnel with a immersion heater, with a beaker underneath (this is the control - like the thing you are using to test against). You set up the same exact thing in another funnel, but this time you connect one to an electric source. You wait till water starts dropping (due to ice melting). As soon as ice starts melting, turn on the immersion heater. Wait for 15 mins so reasonable amount of water can be collected from melting. Measure the mass of water in both beakers and then find the difference. This difference is equal to mass of ice melted due to immersion heater. Use the formula E=mL to find L (specific latent heat).
Reply 5
specific latent heat is the energy needed to change the state. so use equation Q=ml with l being the specific latent heat of the object. for a experiment you will need to know the amount of energy needed to evaporate a given mass of water ,then rearrange the equation to give Q/m=l which gives the specific latent heat of evaporisation for water

Original post by AlishaWhite
Can someone explain an experiment to determine the Specific Latent Heat of Vaporisation? Let's say for water.
Original post by yjacobs
specific latent heat is the energy needed to change the state. so use equation Q=ml with l being the specific latent heat of the object. for a experiment you will need to know the amount of energy needed to evaporate a given mass of water ,then rearrange the equation to give Q/m=l which gives the specific latent heat of evaporisation for water

She asked for the experiment?
Original post by neluxsan
You set an ice in a funnel with a immersion heater, with a beaker underneath (this is the control - like the thing you are using to test against). You set up the same exact thing in another funnel, but this time you connect one to an electric source. You wait till water starts dropping (due to ice melting). As soon as ice starts melting, turn on the immersion heater. Wait for 15 mins so reasonable amount of water can be collected from melting. Measure the mass of water in both beakers and then find the difference. This difference is equal to mass of ice melted due to immersion heater. Use the formula E=mL to find L (specific latent heat).

Thx. That is for fusion right? What would an experiment for vaporisation be?
Reply 8
Original post by AlishaWhite
Thx. That is for fusion right? What would an experiment for vaporisation be?

fusion and evaporisation is the same thing experiment/ constant wise, its just a different state change- i think. also the experiment would be to heat up liquid water until its fully evaporated. then somehow find out how much energy was used and use the equation before
Original post by AlishaWhite
Thx. That is for fusion right? What would an experiment for vaporisation be?

Yes that is for fusion.

For vaporisation, its harder to explain coz idk the equipment names myself... better to look at a picture whilst reading my explaination

You have this spherical container which has gaps to allow gas to flow through, when you heat up water to steam
You have a condenser to cool down the steam and collect it in a beaker
You have an immersion heater fitted inside the spherical container
So you heat up the water and let gas flow into the condenser which cools down and then collects the water.
Energy to heat is calculated via E=I*V*T
Measure mass of water
Use formula E = M*L to find answer
Reply 10
Original post by yjacobs
fusion and evaporisation is the same thing experiment/ constant wise, its just a different state change- i think. also the experiment would be to heat up liquid water until its fully evaporated. then somehow find out how much energy was used and use the equa


Original post by neluxsan
You set an ice in a funnel with a immersion heater, with a beaker underneath (this is the control - like the thing you are using to test against). You set up the same exact thing in another funnel, but this time you connect one to an electric source. You wait till water starts dropping (due to ice melting). As soon as ice starts melting, turn on the immersion heater. Wait for 15 mins so reasonable amount of water can be collected from melting. Measure the mass of water in both beakers and then find the difference. This difference is equal to mass of ice melted due to immersion heater. Use the formula E=mL to find L (specific latent heat).

in this experiment how do you find out E?
Original post by yjacobs
fusion and evaporisation is the same thing experiment/ constant wise, its just a different state change- i think. also the experiment would be to heat up liquid water until its fully evaporated. then somehow find out how much energy was used and use the equation before

Not same experiment because when you use a funnel to heat up the water, it would escape as gas so you wouldn't be able to collect the water(the 'steam') and measure it.
Reply 12
Original post by neluxsan
Not same experiment because when you use a funnel to heat up the water, it would escape as gas so you wouldn't be able to collect the water(the 'steam') and measure it.

true dat
Original post by yjacobs
in this experiment how do you find out E?


If you have a powr source u can attatch a voltmeter and ammeter to it and then time using a stopwatch.

Energy = power x time

So E = V I t
Original post by neluxsan
Not same experiment because when you use a funnel to heat up the water, it would escape as gas so you wouldn't be able to collect the water(the 'steam') and measure it.

Instead of collecting it, could you not measure change in mass (so mass of water evaporated)??

Then use that mass in E=mL
Reply 15
Original post by AlishaWhite
If you have a powr source u can attatch a voltmeter and ammeter to it and then time using a stopwatch.

Energy = power x time

So E = V I t

k, gl with ur experiment
Lo
Original post by AlishaWhite
Instead of collecting it, could you not measure change in mass (so mass of water evaporated)??

Then use that mass in E=mL

I assume that would work too. Cannot think of any drawbacks atm.

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