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Changes of state , latent heat and heat transfers

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the total amount of heat required to heat a 0.2 kg of ice from a temperature of -20 degrees to a temperature of 30 degrees Celsius

    2. Relevant equations

    Q=mcΔt

    Q=mLf

    Q=mLv


    3. The attempt at a solution

    To solve this they break it down, almost like a chemical equation


    Qtotal= Q warm ice + Q Melt ice + Q Warm water (this is general equation)

    My understanding is that whenever we are dealing with heat transfer and the change in the state of an object(solid to liquid or liquid to gas) , we use Q=mLv (if its evaporating or condensing) or Q=mLf (if its freezing or melting)

    so keeping in mind

    Qtotal= Q warm ice + Q Melt ice + Q Warm water

    = m1c1Δt + mLf + m1c1Δt

    and then we continue to solve this....









    and in another problem...

    the question says


    what mass of steam is produced when 2.4 X 10^6 J of heat is applied to 4kg of water at 20 degrees Celsius?

    For this one the general equation is

    according to the book

    Q total = Q warm water + Q boil (this is the general equation)

    Q total = m1c1Δt + mLf


    Now here is what bothers me......in question one...we start from ice, heat transfer occurs, which is represented by the latent heat equation mLf, and then the ice melts. In total there are 3 parts to the general equation...

    in other words form changes...what was in the form of ice, is now in the form of water...


    the same thing happens... in the second question..the form changes..from water..to steam...yet there are only 2 parts of the general equation there...


    Why? how?

    How can i learn to write these general equations? Is there a rule of thumb or something that can tell me how to structure these general equations...or how many parts to put in them?
     
  2. jcsd
  3. May 14, 2012 #2
    The only thumb rule is to apply common sense. If you are asked to calculate the heat required to heat the water from ice at -100C to water at 200C you just need to think it up in your mind, how the process occurs.

    For this specific example, you can say (in a rather loose manner), that the ice wants to reach 200C. So its on its way heating up to 20, but as it gets heated up, there comes a break point at 00C, which indeed is the melting point. So here, it undergoes a phase change into its liquid form, water. But still, the aim of water formed is to reach 200C, and the heating process continues. All they do is write this in equation form,

    Qtotal= Q warm ice + Q Melt ice + Q Warm water

    Similar logic for water getting converted into steam.
     
  4. May 14, 2012 #3


    If you want the most general way, you could always write out all five Q's from solid to gas and cross out the ones that are zero.

    Q total = Q warm ice + Q melt + Q warm water + Q boil + Q warm steam
     
  5. May 14, 2012 #4


    I GOT IT! i think...

    everytime an object reaches a freezing/melting point or boiling/condensing point...and goes past it or beyond it we add another section to the general equation.

    in the first example....the ice gets to the melting point from -20 to 0 degrees...and CONTINUES past it, continuing to heat up till 30 degrees, thats why they have a 3rd part to that general equation...whereas in the second example...once the water hits 100 degrees...it turns to steam...and the increase in temperature stops....it doesnt go past 100 degrees, because it doesnt need to, the water has already become steam, which is why in the book there are only 2 parts to the general equation..not 3!


    So the key here is the freezing/melting and boiling/evaporating points in the temperature.....Thats how you know how many parts there are to the equation...if the temperature goes past the point...you add another part..
     
  6. May 14, 2012 #5
    Yep! Pretty much how it is :smile:
     
  7. May 14, 2012 #6
    Ok...so here is a 3rd question that seems to be the exception to the rule that i seemed to have established above....


    the question is


    " an ice cube with a mass 50 g is transferred directly from the freezer at the temperature of -9 degrees to a polystyrene cup containing 150 g of water at 95 degrees Celsius. Calculate the final temperature of the water."

    the general equation in the book has 4 parts to it....even though i only see the temperature of the ice cube go above melting/freezing point only ONCE, and it never reaches the boiling(evaporating)/condensing point at all....

    so there should only be 3 parts to the equation....but the book shows 4 parts...what gives?


    the general equation the book shows is:-

    Q total = Q heat ice + Q melt ice + Q heat cold water + Q cool hot water
     
  8. May 14, 2012 #7
    Q total = (Q heat ice + Q melt ice + Q heat cold water) + Q cool hot water

    The stuff in the parentheses is what is happening to the icecube and the melt water from the ice. The last term represents the 95 C water.
     
  9. May 14, 2012 #8
    oh..so the last bit isnt part of the process or reaction...???

    but according the book it is...its used in the calculations...

    (in attachment, you can view it without downloading)
     

    Attached Files:

  10. May 14, 2012 #9
    Yes. You still need it. Just like the problem says, the hot water in the cup will cool down. The final temperature is some equilibrium temperature between the ice melting and heating up and the hot water cooling down.
     
  11. May 14, 2012 #10
    yes but how do you come up with that general equation? What is the thought process or train of thought that leads to....ice cube, dropped in hot water, and 4 terms in a general equation?
     
  12. May 14, 2012 #11
    If you want the most general way, you can write down every single Q and cross out the ones that are zero. The total heat that is flowing is heat flowing into the 50g Ice cube and heat flowing out of the 150 g of water.

    Q total = Q 50 + Q150


    First I will look at the 50 gram hunk of ice.

    Q 50 = Q warm ice + Q melt + Q warm water + Q boil + Q warm steam

    These are all the possible ways that the ice can gain heat. We know that the ice is simply being dropped into hot water, so it will never boil. Because of this, two of the Q's are zero

    Q 50 = Q warm ice + Q melt + Q warm water

    Now let's look at all of the ways that the 150 g of hot water can loose heat.

    Q 150 = Q cool ice + Q freeze + Q cool water + Q boil + Q cool steam

    We know that the icecube is just going to cool down the water and it's not going to get cold enough to freeze, so most of these terms are zero.

    Q 150 = Q cool water

    Combining the heat lost by the 150 g of water and the heat gained by the 50 g of ice, you get.

    Q total = (Q warm ice + Q melt + Q warm water) + Q cool water
     
  13. May 14, 2012 #12
    whoa....that last post really made it make sense.....

    Thanks a lot man :)
     
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