Calculate Latent Heat of Steam to Ice Transformation

  • Thread starter Thread starter Jimsac
  • Start date Start date
  • Tags Tags
    Heat Latent heat
Click For Summary
SUMMARY

The discussion focuses on calculating the heat removal required to transform steam at 110°C into ice at 0°C. The correct formula to use is Q = (mcΔT)_{steam} + (mL_{v})_{steam} + (mcΔT)_{water} + (mL_{f})_{water}, where m represents mass, L_{v} is the latent heat of vaporization, L_{f} is the latent heat of fusion, and c is the specific heat of water. The process involves three phases: condensation of steam to water, cooling of water, and freezing of water to ice. It is crucial to account for the latent heat of condensation and to ensure accurate temperature changes during each phase transition.

PREREQUISITES
  • Understanding of latent heat concepts, specifically latent heat of vaporization and fusion.
  • Knowledge of specific heat capacity of water.
  • Familiarity with phase changes of water (steam, liquid, solid).
  • Basic algebra for manipulating equations and solving for heat transfer.
NEXT STEPS
  • Research the specific heat of water and its values at different temperatures.
  • Study the latent heat of vaporization and fusion for water.
  • Learn about phase change diagrams and their applications in thermodynamics.
  • Explore practical examples of heat transfer calculations in thermodynamic processes.
USEFUL FOR

Students and professionals in thermodynamics, physics enthusiasts, and anyone involved in heat transfer calculations or studying phase changes of substances.

Jimsac
Messages
10
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
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.
 
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:
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?
 
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 celsius, any heat lost is due to condensation. Review my equation above.
 
Thanks so much Sirus! After reveiwing your equation a couple of times I got it.
 

Similar threads

Replies
12
Views
1K
Replies
14
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
654
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K