9. Lake Champlain, bordered by New York, Vermont, and Quebec, has a surface area

Question

9. Lake Champlain, bordered by New York, Vermont, and Quebec, has a surface area of ,km^2 (squared). (a) if it freezes solid in the winter with an average thickness of 0.5m, how much total energy does it take to melt the ice once spring comes? The density of ice is about 0 kg/m^3 and the heat of fusion is 334 kJ/kg. (b) If this energy is provided by sunlight shining on the take at an average rate of 200W/m^2 but with only half the amount absorbed, how long will it take the lake to absorb the energy needed to melt it?

Explanation / Answer

Strategy for part a.

We have to figure out how much energy (heat) we have to put in to melt the ice. To do this we use the equation

Q=ml where Q is heat, m is the mass of the ice and L is the latent heat of fusion.

How do we figure out the mass of the ice? Well, we know the surface area and the thickness. From this we can

calculate the volume, V=Sa * T. We also know the density and since D=m/v, we can solve for mass as m=D*V.

Now

that we have a strategy, let's solve part a.

V= Sa * T= 1,126 km^2 *0.5 meters ( or 5 *10^-4 meters, watch those units!)= .563 km^3 of ice.

Okay now we have the volume, let's solve for mass.

m= D*V, again watch the units .563 km^3 = 563 000 000 m^3 * 920 kg/ m^3= 5.152 * 10^11kg of ice

Now we have the mass, let's plug that into Q=ml, Q= 5.152 * 10^11kg * 334KJ/Kg= 1.72*10^14 KJ.
We're done!

Now let's attack part B.

Here's our stratey. We need to figure out how long it will take for the lake to absorb

1.72*10^14 KJ of energy. Power (watts) tells us the amount of energy per time that we get from the sun and

intensity (power/area) tells us how much of that energy the lake is receiving. So, if we multiply the intensity * the

surface area * the energy we need, we should get the time it will take to melt. Be extremely careful of units. In this

case everything needs to be in meters, seconds, and joules

Here's the math
(1 second/200 J *meter^2) X (1.126 *10^9 meters^2) (1.72 X10^14 Kilojoules) X (1000 Joules/ KJ) X 50%

= 4.84*10^23

Related Questions

Navigate

• Browse (All)
• Browse 9
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.