Latent heat, steam and ice question

In summary, the problem involves adding 14g of steam at 100C to 47g of ice at 0.0 degrees C. The solution requires considering the various processes that affect heat, such as phase changes and specific heat capacity. After making some corrections to the equations, the final temperature is found to be 115.77C, which is incorrect. After further calculations, it is determined that the solution involves calculating the overall energy released by the condensation of the steam and using this to raise the temperature of the remaining water and steam, resulting in a final temperature of around 70C.
  • #1
TheStebes
11
0
[SOLVED] Latent heat, steam and ice question

Homework Statement



14g of steam at 100C is added to 47g of ice at 0.0 degrees C.
a) Find the amount of ice melted and the final temperature.

Homework Equations


Q=mL
Q=mc[tex]\Delta[/tex]T
Q(hot)=-Q(cold)

The Attempt at a Solution


Ordinarily, in such a problem, you would have to consider the various processes which affect Q(hot) and Q(cold). However, in this problem, the starting temperatures are 100 and 0 C, so the heat or energy goes directly to the phase change, correct?

m[tex]_{steam}[/tex]=.014kg
m[tex]_{ice}[/tex]=.047kg
L[tex]_{vaporization}[/tex]=2.26*10^6
L[tex]_{fusion}[/tex]=3.33*10^5
c[tex]_{water}[/tex]=4186

Q(hot)= -m[tex]_{s}[/tex]*L[tex]_{v}[/tex] + m[tex]_{s}[/tex]*c[tex]_{w}[/tex]*T[tex]_{f}[/tex]

Q(cold)= m[tex]_{i}[/tex]*L[tex]_{f}[/tex] + m[tex]_{i}[/tex]*c[tex]_{w}[/tex]*T[tex]_{f}[/tex]

set
Q(hot)=-Q(cold)

solving for T[tex]_{f}[/tex], I get 62.62 degrees C.

Obviously I'm misusing or neglecting one or more equations in the Q calculations, as my solution does not even solve for the mass of the water. I'm thoroughly confused as to which equation to use with which variables. (between latent heat and the specific heat capacity equation.)
 
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  • #2
I'm sorry, I meant to use subscripts not superscripts!
 
  • #3
did I leave anything out? Not show enough work? My T(final) seems reasonable, but was not the correct answer.
 
  • #4
After the vapour condenses it is at 100C. To cool down to [itex] T_f [/itex] it has to lose more heat. the second term of Q(hot) should also be a negative number.
 
  • #5
with that in mind, solving for T yields 342.44, and unless I messed up something with units, that is already in Celsius. I would think the value would be between 0 and 100...
 
  • #6
TheStebes said:
with that in mind, solving for T yields 342.44, and unless I messed up something with units, that is already in Celsius. I would think the value would be between 0 and 100...

what is the heat loss if 14g of water at 100C cools to T_f ? put that in the equation for Q(hot)
 
  • #7
I realized I messed up the signs of Q(cold) in the equation when trying your first suggestion of making the second term of Q(hot) negative. So my answer of 342 was off. However even after fixing that issue, I get 115.8.

So...perhaps I'm misunderstanding, but as far as I can tell, I have incorporated the cooling of the water into the expression for Q.

Q(hot) = latent heat of condensing vapor + cooling of water to T(final).
Q(cold) = latent heat of melting ice + heating of water to T(final).

setting Q(hot) = -Q(cold)...

-m(steam)L(v)+ -m(steam)c(water)T(f) = -[m(ice)L(f) + m(ice)c(water)T(f)]

solving for T(f) yields 115.77, which as far as I can tell, is still wrong.

Perhaps I misunderstood your last comment?

Thanks for the help so far,
Scott
 
  • #8
in the specific heat equation for the water coming from steam, instead should it be:

m[itex]_{s}[/itex]c[itex]_{w}[/itex][T[itex]_{f}[/itex]-100]

s=steam
w=water
 
Last edited:
  • #9
Nothing seems to be working. I've tried every combination of signs (pos/neg) I can think of. Clearly I'm missing something from one of the equations. Even if I solve for the correct temperature, that is only half the solution.

Any other suggestions?
 
  • #10
I seem to have figured it out. Basically, you calculate the overall energy (heat) released by the condensation of the steam. This is more than enough to melt all 47g of ice. So now you have 47g of water at 0C, a fraction of the water from steam at 100C, and the remaining steam itself. Using the remaining energy from the steam, I calculated much this could raise the temperature of the 0C water, then did a weighted average between this value and the water still at 100C.
 

1. What is latent heat?

Latent heat is the energy required to change the state of a substance without changing its temperature. This energy is used to break or form the bonds between molecules, resulting in a change from solid to liquid or liquid to gas, or vice versa.

2. How is steam formed?

Steam is formed when water is heated to its boiling point and undergoes a phase change from liquid to gas. This process requires a significant amount of latent heat to break the bonds between water molecules.

3. What is the latent heat of vaporization?

The latent heat of vaporization is the amount of energy required to change a substance from liquid to gas at its boiling point. For water, this value is 2260 joules per gram.

4. How does latent heat affect the melting and freezing of ice?

When ice melts, it absorbs energy from its surroundings in order to break the bonds between water molecules and change from a solid to a liquid. This energy is known as latent heat of fusion. Conversely, when liquid water freezes, it releases energy into its surroundings as it forms bonds and becomes a solid again.

5. How is latent heat related to the heat of fusion and heat of vaporization?

The heat of fusion and heat of vaporization are specific types of latent heat. The heat of fusion is the latent heat required to change a substance from solid to liquid, while the heat of vaporization is the latent heat required to change a substance from liquid to gas. Both values depend on the substance and its surrounding conditions.

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