How Do You Calculate Heat Released When Steam Turns to Ice?

[lolphysics3]

Homework Statement

How much heat is released freezing 75 grams of steam at 110 degrees celcius into ice at -5.1 degree celcius

Homework Equations

The Attempt at a Solution

I tried to do this problem but got an answer like -17101.6J. I don't need you guys to give me an answer but can you suggest how to setup a problem like this? I know Qlost+Qgained=0...so would the 2 phase changes (steam to liquid, liquid to solid) =0?

 

[Delphi51]

Several parts:
- cooling the steam
- condensing the steam to water
- cooling the water to freezing point
- freezing
- cooling the ice

 

[lolphysics3]

So would I have (Msteam*Lheatsteam)+(Mwater*Lheatwater)+(Mice*Lheatice)+(Mice*Sice+change in temp ice)=0?

 

[Delphi51]

(Mice*Sice+change in temp ice)=0?

Certainly not equal to zero - equal to the heat released that you are trying to find.

I don't know what "Sice" means. I would say
M*C*ΔT for each of the three coolings, where M is the mass, C the heat capacity per kg per degree and ΔT the change in temperature.

 

[rwisz]

To solve this problem, we can use the equation Q = mCΔT, where Q is the heat change, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature. Since we are going from steam to ice, we will need to account for the two phase changes (steam to liquid, liquid to solid) as well as the change in temperature.

First, we need to calculate the heat lost by the steam as it cools from 110°C to 100°C (boiling point of water). We can use the specific heat capacity of steam, which is 2.01 J/g°C. So the heat lost in this step would be:

Qlost = (75 g)(2.01 J/g°C)(110°C - 100°C) = 1507.5 J

Next, we need to account for the heat lost during the phase change from steam to liquid. We can use the latent heat of vaporization of water, which is 2257 J/g. So the heat lost in this step would be:

Qlost = (75 g)(2257 J/g) = 169275 J

Now we have reached the boiling point of water, and the steam will start to condense into liquid water. We need to calculate the heat lost as the temperature of the liquid water cools from 100°C to 0°C. We can use the specific heat capacity of water, which is 4.18 J/g°C. So the heat lost in this step would be:

Qlost = (75 g)(4.18 J/g°C)(100°C - 0°C) = 31125 J

Finally, we need to account for the heat lost during the phase change from liquid to solid. We can use the latent heat of fusion of water, which is 334 J/g. So the heat lost in this step would be:

Qlost = (75 g)(334 J/g) = 25050 J

Now we have reached the freezing point of water, and the liquid water will start to freeze into ice. We need to calculate the heat lost as the temperature of the ice cools from 0°C to -5.1°C. We can use the specific heat capacity of ice, which is 2.03 J/g°C. So the heat lost in this step would be:

Qlost = (75 g)(2.03 J/g°C)(0