Calculating Heat Release: Freezing a 8.5 km^2 Lake to 1.0m Depth

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The discussion revolves around calculating the heat released when a lake with a surface area of 8.5 km² freezes to a depth of 1.0 m, starting with water at 0 degrees Celsius. The correct answer is noted as 6.2 x 10^11 Kcal. Participants emphasize that the problem is straightforward, as there is no temperature change involved due to the water being at 0°C. The key steps include calculating the volume of ice formed (area multiplied by thickness) and applying the heat of fusion of ice, which is expressed in Kcals per mole. One contributor mentions using the number of moles of water frozen and the heat of fusion in kJ/mol for the calculation. The discussion concludes with a note of confusion regarding the supplied answer, suggesting that it may be incorrect.
jimmyche
Can anyone help me with the following question.

How much heat is released when a lake of surface area 8.5 km^2 freezes to a depth of 1.0m? assume the water is initially at 0 degree C. The answer is 6.2 X10^11 Kcal.

Thanks.

any help is approciated.
 
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Look up Newton's law of cooling, it should help you to figure that one out.
 
It is really much simpler then Newtons law of cooling. Since the water is at 0C already there is no temperature change involved. You know the volumn of Ice forming (area x thickness) and you should know about the heat of fusion of Ice, this is given as some number of Kcals per mole of water, simple multiplication.

If I told you any more I would be doing the problem for you.
 
Since I'm a chemist, i would find out the number of mol's of water you have frozen and then use the kj/mol of fusion. Thus giving it to you. but you physics people might have an easier way :-)

Pete
 
got it.
thanks everyone.

is just that the supplied answer is somehow wrong which confuse me.
 
Thread 'Thermo Hydrodynamic Effect'

Thread 'Thermo Hydrodynamic Effect'

Vídeo: The footage was filmed in real time. The rotor takes advantage of the thermal agitation of the water. The agitation is uniform, so the resultant is zero. When the aluminum cylinders containing frozen water are immersed in the water, about 30% of their surface is in contact with the water, and the rest is thermally insulated by styrofoam. This creates an imbalance in the agitation: the cold side of the water "shrinks," so that the hot side pushes the cylinders toward the cold...
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