# Latent heat

1. Jan 2, 2005

### Jimsac

I have to change steam at 110C and end up with water frozen into ice at 0.c . How much heat will be removed?

How do I go about changing this to steam, liquid, solid?
Do I use mLf+mcdeltaT each time?

2. Jan 2, 2005

### Q_Goest

There's a latent heat which is removed to change the water vapor (steam) into liquid. There is a heat which results in the water cooling from the boiling point to the freezing line. And finally there is a latent heat which is removed to change the water into ice.

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. Jan 2, 2005

### HallsofIvy

Staff Emeritus
One of the things you don't mention is how much steam you are condensing.

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. Jan 2, 2005

### Sirus

$$Q=(mc\Delta{T})_{\mbox{steam}}+(mL_{\mbox{v}})_{\mbox{steam}}+(mc\Delta{T})_{\mbox{water}}+(mL_{\mbox{f}})_{\mbox{water}}$$

where $L_{\mbox{v}}$ is latent heat of vaporization and $L_{\mbox{f}}$ 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.

Last edited: Jan 2, 2005
5. Jan 2, 2005

### Jimsac

does this look right?

(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. Jan 2, 2005

### Sirus

I do not understand your work (esp. second line). Please include units for all values and explain what you are doing. You should not have a $\Delta{T}$ of 110 degrees - 0 degrees because there is a phase change in between; when steam cools to 100 degrees Celcius, any heat lost is due to condensation. Review my equation above.

7. Jan 4, 2005

### Jimsac

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