username5838300
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When finding the latent heat of vaporisation of water, I have to use the 2 sets of data 'to allow for heat loss'. What does 'to allow for heat loss' mean? It's so confusing...
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RogerOxon
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(Original post by hiyeonnn)
When finding the latent heat of vaporisation of water, I have to use the 2 sets of data 'to allow for heat loss'. What does 'to allow for heat loss' mean? It's so confusing...
Not all the heat energy, supplied by the heater, is used to produce steam. A certain amount will be required to keep the water at 100C - heat is lost to the surroundings. That's why there are two data point. How can you see what the heat loss is?
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username5838300
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(Original post by RogerOxon)
Not all the heat energy, supplied by the heater, is used to produce steam. A certain amount will be required to keep the water at 100C - heat is lost to the surroundings. That's why there are two data point. How can you see what the heat loss is?
Well the mark scheme states that the energy loss is (11.5 × 5.2)J, which is what I don't get.
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(Original post by hiyeonnn)
Well the mark scheme states that the energy loss is (11.5 × 5.2)J, which is what I don't get.
If you use a single measurement, you don't know what the heat energy lost to the surroundings is. If you assume that it is a constant (the water is always held at 100C), then taking the difference between two measurents will remove it.
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RogerOxon
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(Original post by hiyeonnn)
Well the mark scheme states that the energy loss is (11.5 × 5.2)J, which is what I don't get.
No, it doesn't. It's taking the difference in input power / energy, and equating that to the difference in latent heat of evaporation. It doesn't show the loss - h or Q in its notation.

It's not explained in the mark scheme - it is not using the first equations, but the difference between their values for the two sets of measurements, to cancel out h/Q.

You know that there is a constant heat loss, but if you take the extra energy required to vapourise another 9.1 - 5 grams, then you have removed the constant heat loss.
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username5838300
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(Original post by RogerOxon)
No, it doesn't. It's taking the difference in input power / energy, and equating that to the difference in latent heat of evaporation. It doesn't show the loss - h or Q in its notation.

It's not explained in the mark scheme - it is not using the first equations, but the difference between their values for the two sets of measurements, to cancel out h/Q.

You know that there is a constant heat loss, but if you take the extra energy required to vapourise another 9.1 - 5 grams, then you have removed the constant heat loss.
Thank you so much! I finally get it.
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RogerOxon
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(Original post by hiyeonnn)
Thank you so much! I finally get it.
Good. The mark scheme was not good on this, so it's understandable that it confused you.
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