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Latent heat problem

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data

    When spring finally arrives in the mountains, the snow pack may be two meters deep, composed of 50 % ice and 50 % air. Direct sunlight provides about 1000 wats /m^2 to earth's surface, but the snow might reflect 90 % of this energy. Estimate how many weeks the snow pack should last, if direct solar radiation is the only source of energy.


    2. Relevant equations

    Possible equations: Q=L/m, Q=C*delta(T)

    3. The attempt at a solution

    L_water=333 J/g, L_boiling water=2260 J/g

    Not really sure how to start this problem. I know I am supposed to derived an equation for time,

    I would probably calculate the surface Area X Power/area to get the power. Then I would have to calculate the energy of the direct sun then I would divide the power by energy to ge the time.
     
  2. jcsd
  3. Feb 5, 2009 #2

    Gokul43201

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    You have written down 2 relevant equations. The first one has an error in it. Find the error and fix it.

    Now what do these two equations represent? Which one of them applies to this problem, or do you need both of them? Describe the physical process that is occurring in the situation described by the problem.
     
  4. Feb 5, 2009 #3
    You will need to figure out how much energy per unit area is needed to melt ice with the given depth. You also need to figure out the rate at which energy is being absorbed by the ice.
     
  5. Feb 6, 2009 #4
    I found the error. Q=Lm , not Q=L/m. I think either equations would be useful. I will used the latent heat equation. They both represent the same thing: which is how much heat is needed to melt or boil a substance. since 90 percent of the energy is reflected, 1000 watts/m^2 becomes 100 watts/m^2. I don't need to determined the latent heat for air since air isn't the substance that is undergoing a phase transition, so I just need to determined Q_ice. m_water=rho*V_water. Estimate the Volume of water to be :V_water=(2m)^3=8m^3. ==> m_water=(1000 grams/m^3)*(8m^3)=8000 grams. Therefore, Q_water=(333J/g)(.5)(8000 grams)= 1332000 joules I need to convert P/A to P since I need P to find the time it takes for ice to melt to water. What do you think a good estimate of the radius of a pack of ice will be? I said r=5 m. Therefore the surface area of water is: SA_water=4*pi*(5)^2=314 m^2. Therefore P=(P/A)*SA_water=(100 watts/m^2)*(314 m^2) =31400 watts. I got my Q, which is my energy and I got my Power. Now I can find the time t. Q/t=P ==> t=Q/P= 42.42 seconds which is way too short of a time for ice to melt. Where did I go wrong?
     
  6. Feb 6, 2009 #5
    The density of water is 1g/cm^3, which works out to 1000 kg/m^3. You wrote the density of water as 1000 g/m^3, which is three orders of magnitude too low.
     
  7. Feb 7, 2009 #6
    But everything else I did was correct? I just had the wrong density?
     
  8. Feb 10, 2009 #7

    Gokul43201

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    1. You are only given the depth = 2m. So, when you estimate V=(2m)^3, you are implicitly assuming that the surface area is A = (2m)^2 = 4m^2

    2. Your density is off, and you are using the density for the wrong material. If you are melting ice, it is the mass of ice that appears in the equation, not the mass of water. Fortunately, ice has almost exactly the same density as water, but you need to get this number right.

    3. Where did the "(.5)" come from??

    4. See point #1. You have already assumed an area. You need to use that same area.

    5. See points #1 through #4.

    In general, it is good practice to solve the problem using only symbols to represent various quantities involved (area, depth, latent heat, density, solar power, reflectance, time, etc.), and arrive at a final expression involving the symbols only. Then in the final step, you plug in the values for the quantities to arrive at a numerical solution. If you do this, you will see that in your final expression, the symbol for the area of the pack cancels off and does not appear in the answer.
     
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