Solving Heat Transfer Rate for 10 mm Ice Layer at -5°C

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In summary, the conversation discusses the calculation of the rate of increase in the thickness of ice on a frozen lake when cool air blows across its surface. The solution involves considering the heat transfer coefficient and heat flux, and ultimately results in a thickness change of 2.799*10^-6 meters per second. The logical thinking behind this solution is also explained.
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debwaldy
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Homework Statement


hi,i was wondering if someone could tell me if the following solution makes sense,ithink it does but not sure why?


cool air blowing across the surface of a frozen lake keeps the top surface of the ice at a temperature of -5degrees celsius.what is the rate of increase in the thickness of the ice layer when the ice is 10 mm thick?

Homework Equations



the density,thermal conductivity and specific latent heat of fusion of ice are 920 kg m^-3, 1.7 J m^-1 K^-1 s^-1 and 3.3*10^5 J kg^-1 respectively

The Attempt at a Solution


so i said that:

the heat transfer coefficient = thermal conductivity/thickness of material
i.e heat transfer coefficient= 1.7/0.01= 1.7*10^2 J K^-1 s^-1

then i said :
heat flux=(1.7*10^2) *5 = 8.5*10^2

then:
8.5*10^2/920 = 9.23913*10^-1

and so:
9.23913*10^-1/3.3*10^5= 2.799* 10^-6 ms^-1

i know the answer is right but i don't quite understand the thinking behind it.could anyone explain it to me?
any help would be much appreciated:biggrin:
 
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  • #2
You seem to have done the right sums, but not in a logical order.
Your heat transfer coeff. and heat flux are OK.

Now think about what happens to 1 square meter of lake surface in 1 second.

850 J of heat flows out of the lake. That freezes 850/3.3*10^5 = 2.575*10^-3 Kg of ice.

The volume of the ice is 2.575*10^-3/920 = 2.799*10^-6 m^3.

We were considering 1 square meter of area, so 2.799*10^-6 m^3 is a block of size 1m * 1m * 2.799*10^-6 m.

The time was 1 second, so the thickness changes at 2.799*10^-6 ms-1.
 
  • #3
thanks a million,that makes sense to me now alright...wasnt really thinking bout it in a logical manner
thanks for clearing that up!
:biggrin:
 

FAQ: Solving Heat Transfer Rate for 10 mm Ice Layer at -5°C

1. How is heat transfer rate calculated for a 10 mm ice layer at -5°C?

The heat transfer rate for a 10 mm ice layer at -5°C can be calculated using the Fourier's law of heat conduction, which states that the heat transfer rate is directly proportional to the temperature difference and the thickness of the material, and inversely proportional to the thermal conductivity of the material.

2. What is the thermal conductivity of ice at -5°C?

The thermal conductivity of ice at -5°C is approximately 2.22 W/mK. However, this value can vary depending on the purity and density of the ice.

3. How does the thickness of the ice layer affect the heat transfer rate?

The thickness of the ice layer directly affects the heat transfer rate, as stated by Fourier's law. A thicker ice layer would result in a slower heat transfer rate, while a thinner ice layer would result in a faster heat transfer rate.

4. Can the temperature difference between the ice layer and the surrounding environment affect the heat transfer rate?

Yes, the temperature difference between the ice layer and the surrounding environment has a direct impact on the heat transfer rate. A larger temperature difference would result in a faster heat transfer rate, while a smaller temperature difference would result in a slower heat transfer rate.

5. How can the heat transfer rate be increased for a 10 mm ice layer at -5°C?

The heat transfer rate for a 10 mm ice layer at -5°C can be increased by reducing the thickness of the ice layer, increasing the temperature difference between the ice layer and the surrounding environment, or using a material with a higher thermal conductivity for the ice layer.

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