Thermodynamics: Final temp of Ice + steam?

[Tabatron]

Homework Statement

In an insulated container, .50kg of steam at 140C is mixed with 2.0kg of ice, at -20C. What is the final temp inside the container if heat exchanges with the container are ignored?

This is a multiple choice question and the answer is 60C.

Homework Equations

mC(Tf-Ti)
Lf
Lv

The Attempt at a Solution

I'm stuck on how to figure out what mass of the ice and steam will turn into water, do I assume that all turns to water? I'm also confused on where to go after that.

Qsteam = .5(2010)(100-140) + m(22.6 x 10^5) + m(4186)(Tf-100)
Qice = 2(2090)(0-(-20)) + m(33.5 x 10^4) + m(4186)(Tf-0)

As for the weird Lf and Lv values, their in J/Kg and I'm just using the numbers given to me from my teachers formula chart.

EDIT: Ok, so I assumed that all turns to water and got an answer of Tf = -163, which I remember my teacher saying that this means not all of the ice melts. Is this correct?

Any help is appreciated.

 

[Mindscrape]

Okay, good job on factoring in the changes of state. You seem to be having trouble with the units of Lf and Lv? Anyway, if you do have those correct, you are basically there. You just have to realize that the two heats must be equal because the container is thermally isolated, the heat can only exchange within the system.

 

[Tabatron]

The Lf and Lv values I obtained of my formula chart.

I'm still confused though, how do I solve for how much mass of ice will melt?

 

[Mindscrape]

Since you've already assumed that the two states are going to both turn into water, which is smart of course, then just use the respective masses and assume they all turn to water (that no mass went into energy).

 

[rwisz]

I would approach this problem by first understanding the principles of thermodynamics. In this case, we are dealing with a closed system, meaning that there is no exchange of matter with the surroundings. The only exchange happening is through energy in the form of heat.

To find the final temperature, we can use the principle of conservation of energy, which states that the total energy of a closed system remains constant. This means that the total energy before mixing must be equal to the total energy after mixing.

For the steam, we can use the equation Q = mCΔT, where Q is the heat exchanged, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature. We can calculate the heat exchanged by the steam as Qsteam = 0.5(2010)(100-140) = -40200 J (note the negative sign because the steam is losing heat).

For the ice, we can use the equation Q = mL, where L is the latent heat of fusion. We know that all of the ice will melt, so we can calculate the heat exchanged by the ice as Qice = 2(2090)(33.5 x 10^4) = 139.3 x 10^6 J.

Now, we can set these two equations equal to each other and solve for the final temperature (Tf).

-40200 J = 139.3 x 10^6 J + m(4186)(Tf-100)

Solving for Tf, we get Tf = 60.3 degrees Celsius.

So, the final temperature inside the container will be 60.3 degrees Celsius. This means that not all of the ice will melt, as we can see from the negative sign in front of the heat exchanged by the steam. This indicates that the steam will lose more heat than the ice can absorb, resulting in some of the ice remaining in solid form.

In conclusion, as a scientist, I would approach this problem by applying the principles of thermodynamics and using equations to calculate the final temperature. I would also consider the physical properties of the substances involved and use them to make logical conclusions about the final state of the system.