Energy math question - Energy to melt lake ice sheet

  • #1
Energy math question -- Energy to melt lake ice sheet

Homework Statement



Lake Champlain bordered by New York, Vermont and Quebec has a surface area of 1, 126 km^2. If it freezes solid in the winter with an average ice thickness of .5 m, how much total energy does it take ti melt the ice once spring comes? The density of ice is about 920 kg/m^3 and the heat of fusion is 334 kJ/kg.

Homework Equations





The Attempt at a Solution



I don't even know what formula to use for this problem and am very confused. If someone could do this out for me that would be great!
 

Answers and Replies

  • #2
rock.freak667
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If you have the surface area and the thickness, then what is the lakes' volume and hence mass?

What formula do you have to calculate latent heat?
 
  • #3
I am still not sure what formula to use. Q=mL doesn't help I don't think?
 
  • #4
rock.freak667
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Have you learned about specific heat and latent heat or at least what "latent heat of fusion" means?
 
  • #6
rock.freak667
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Can you please give me the formula to use?

The formula you posted would be the one to use, but it seems that you do not know why one would have to use that formula. That's why I asked if you knew what the terms meant.
 
  • #7
I have not learned about latent heat! I am not sure how to start this problem.
 
  • #8
SteamKing
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How many cubic meters of ice are there in Lake Champlain?

What is the mass of this volume of ice, given the density of ice is 920 kg/m^3?

If it takes 334 kJ/kg to melt 1 kg of ice, how many kJ does it take to melt all of the ice in Lake Champlain?

Hint: examine the units at each step of the calculations above.
 
  • #9
nasu
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I have not learned about latent heat! I am not sure how to start this problem.

Maybe it was called "heat of fusion" ?
Not all textbooks use the term "latent heat". (in US and Canada).
It must be some name for that L in the formula you wrote.
 
  • #10
Maybe it was called "heat of fusion" ?
Not all textbooks use the term "latent heat". (in US and Canada).
It must be some name for that L in the formula you wrote.


So I would use the heat of fusion number in the problem? But I don't know Q or m right?
 
  • #11
nasu
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Q is what you are supposed to calculate.
L can be found in tables of heat of fusion. See in the textbook or online.
The mass you will calculate from the given information (area, thickness, density) about the ice layer.
 
  • #12
I think my trouble then is calculating the mass. Can you help?
 
  • #13
Energy Problem

Homework Statement



Lake Champlain bordered by New York, Vermont and Quebec has a surface area of 1, 126 km^2. If it freezes solid in the winter with an average ice thickness of .5 m, how much total energy does it take ti melt the ice once spring comes? The density of ice is about 920 kg/m^3 and the heat of fusion is 334 kJ/kg.


Homework Equations



Q= mL

3. The Attempt at a Solution [/b

When I looked at this problem, I thought to use the equation Q=mL. L is 334 kJ/kg. However, since I do not have the mass and am trying to solve for Q I am confused.

My guess would be that in order to find the mass you need to use the density. Although I am not sure of this.

Can someone please explain to me how I can get the mass?
 
  • #14
Integral
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You are given surface area and ice thickness.
 
  • #15
You are given surface area and ice thickness.

What am I supposed to do with the surface area and ice thickness?
 
  • #16
berkeman
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Homework Statement



Lake Champlain bordered by New York, Vermont and Quebec has a surface area of 1, 126 km^2. If it freezes solid in the winter with an average ice thickness of .5 m, how much total energy does it take ti melt the ice once spring comes? The density of ice is about 920 kg/m^3 and the heat of fusion is 334 kJ/kg.

Homework Equations





The Attempt at a Solution



I don't even know what formula to use for this problem and am very confused. If someone could do this out for me that would be great!

Homework Statement



Lake Champlain bordered by New York, Vermont and Quebec has a surface area of 1, 126 km^2. If it freezes solid in the winter with an average ice thickness of .5 m, how much total energy does it take ti melt the ice once spring comes? The density of ice is about 920 kg/m^3 and the heat of fusion is 334 kJ/kg.


Homework Equations



Q= mL

3. The Attempt at a Solution [/b

When I looked at this problem, I thought to use the equation Q=mL. L is 334 kJ/kg. However, since I do not have the mass and am trying to solve for Q I am confused.

My guess would be that in order to find the mass you need to use the density. Although I am not sure of this.

Can someone please explain to me how I can get the mass?


Please do not make multiple threads with the same question. Check your PMs...
 
  • #17
21,053
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Volume of ice = (Area of Ice) x (Thickness of Ice)

Mass of Ice = (Volume of Ice) x (Density of Ice)
 
  • #18
Using the above post I was able to get 173 J.

I now need to find if this energy provided by sunlight shining on the lake at an average rate of 200 W/m^2 but with only half the amount absorbed, how long will it take the lake to absorb the energy needed to melt it?

I know i should first find half of 173 which is 86.5 J.

I am trying to convert 200 W to J but don't know how to do this because I am unsure of the time here.

I know you need to be in seconds, but for this problem from winter to spring I am not sure what to use.
 
  • #19
rock.freak667
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How did you get such a low value of 173 J? The surface area is 1126 km^2 so the volume and hence mass would be in a different order.

If 200 W/m^2 is being directed onto the lake and only half is absorbed then 100 W/m^2 is absorbed.

So for the total surface area how much Watts is the ice absorbing?

If 1 W = 1 J/s then how much is the above quantity in J/s?
 
  • #20
I'm not sure why this is so low? But it is also the answer in the back of the book.

I am still confused on how to convert this to joules.
 
  • #21
rock.freak667
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I'm not sure why this is so low? But it is also the answer in the back of the book.

I am still confused on how to convert this to joules.

Watt is a unit of power so that Power = change in energy/time.

EDIT: could you post how you got 173 J?
 

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