- #1

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How do I go about changing this to steam, liquid, solid?

Do I use mLf+mcdeltaT each time?

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- Thread starter Jimsac
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- #1

- 10

- 0

How do I go about changing this to steam, liquid, solid?

Do I use mLf+mcdeltaT each time?

- #2

Q_Goest

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Note these are changes to the water's internal energy, not enthalpy.

I'm not sure what the variables are in your equation, so I'll have to guess that:

m = mass of water

Lf = latent heat of freezing?

c = specific heat of liquid water

deltaT is temperature drop from 110 to 0 C.

Ok, if so that looks right, except you need to add to that the latent heat of condensing.

- #3

HallsofIvy

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The heat extracted will be the mass of the steam times the latent heat of boiling plus the mass of the water times the specific heat of water times the change in temperature (100 degrees) plus the mass of the water times the latent heat of freezing.

- #4

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[tex]Q=(mc\Delta{T})_{\mbox{steam}}+(mL_{\mbox{v}})_{\mbox{steam}}+(mc\Delta{T})_{\mbox{water}}+(mL_{\mbox{f}})_{\mbox{water}}[/tex]

where [itex]L_{\mbox{v}}[/itex] is latent heat of vaporization and [itex]L_{\mbox{f}}[/itex] is latent heat of fusion. Make sure you use the correct values for c in each part.

HallsofIvy, don't forget that the steam is at 110 degrees, not 100.

where [itex]L_{\mbox{v}}[/itex] is latent heat of vaporization and [itex]L_{\mbox{f}}[/itex] is latent heat of fusion. Make sure you use the correct values for c in each part.

HallsofIvy, don't forget that the steam is at 110 degrees, not 100.

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- #5

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(200g)(540) + (200g)(.50)(110degreesC-0degressC)

(108kg) + (11)= 119kg

Is this the problem from steam to ice? Do I have to now bring it to another phase?

- #6

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- #7

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Thanks so much Sirus! After reveiwing your equation a couple of times I got it.

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