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

In summary, the question is asking for the amount of heat released when a lake with a surface area of 8.5 km^2 freezes to a depth of 1.0m, assuming the water is initially at 0 degrees C. The answer is 6.2 X10^11 Kcal and it can be calculated by finding the volume of ice formed and using the heat of fusion of ice. Some respondents suggest using Newton's law of cooling, but others argue that it is simpler to calculate the heat of fusion. It is also noted that the given answer may be incorrect.
  • #1
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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  • #2
Look up Newton's law of cooling, it should help you to figure that one out.
 
  • #3
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.
 
  • #4
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
 
  • #5
got it.
thanks everyone.

is just that the supplied answer is somehow wrong which confuse me.
 

1. How do you calculate the amount of heat needed to freeze a 8.5 km^2 lake to a depth of 1.0m?

To calculate the amount of heat needed, you would use the formula Q = m x Cp x ΔT, where Q is the heat energy, m is the mass of the lake, Cp is the specific heat capacity of water, and ΔT is the change in temperature (in this case, the difference between the initial temperature of the lake and the freezing point of water).

2. What is the specific heat capacity of water?

The specific heat capacity of water is 4.186 joules per gram per degree Celsius. This means that it takes 4.186 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

3. How would you determine the mass of the lake?

To determine the mass of the lake, you would need to know the density of water and the volume of the lake. The density of water is 1000 kg/m^3, so you can multiply the volume of the lake (8.5 km^2 x 1.0m = 8.5 km^3) by the density to get the mass in kilograms.

4. What is the freezing point of water?

The freezing point of water is 0 degrees Celsius or 32 degrees Fahrenheit. This is the temperature at which water changes from a liquid to a solid state.

5. Are there any other factors that could affect the calculation of heat release for freezing a lake?

Yes, there are other factors that could affect the calculation, such as the initial temperature of the lake, the rate at which the lake is frozen, and any external sources of heat or cooling. These factors should be taken into consideration when making the calculation.

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