Using thermal conductivity to melt ice

In summary, the ice cube at 0.00 °C with a side length of 14.9 cm and a density of 0.917 g/cm3 will take approximately 232 seconds to melt when placed on top of a copper block with a side length of 14.9 cm, a height of 18.1 cm, and a thermal conductivity of 385 W/(m*K). This is assuming that only the part of the ice cube in contact with the copper block melts and that the initial temperature profile of the block is linear with the top and bottom at 0 and 92.5 °C respectively. Slight variations in the thermal conductivity value may result in a slightly different answer, but any answer within the
  • #1
castrodisastro
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Homework Statement


An ice cube at 0.00 °C measures 14.9 cm on a side. It sits on top of a copper block with a square cross section 14.9 cm on a side and a height of 18.1 cm. The bottom of the copper block is in thermal contact with a large pool of water at 92.5 °C. How long does it take the ice cube to melt? Assume that only the part in contact with the copper liquefies, that is, the cube gets shorter as it melts. The density of ice is 0.917 g/cm3.

Tice = 273 K
Twater = 365.5 K
LHeat of fusion = 334 kJ/kg
Vice = (14.9cm)3 = 3,307.95 cm3
h = 0.181 m
κCu = Thermal conductivity of Copper = 385 W/(m*K)
A = face area of Cu = (0.149m)2 = 0.0222 m2


Homework Equations


ρ = m/V
Q = mLHeat of fusion
PCond = Q/t = Aκ(ΔT/L)


The Attempt at a Solution


I did this is 3 steps.

1)
Used density to obtain the volume of the ice cube

ρi = mi/Vi

mi = ρiVi

mi = (0.917 g/cm3)(3,307.95 cm3)

mi = 3,033.39 g

2)
Used the mass to obtain amount of energy required to melt the ice, Qi.

Qi = miLHeat of fusion

Qi = (3,033.39g)(334 J/g)

Qi = 1,013,152.3 J

3)
Used Qi and the conduction rate equation to calculate the time t.

PCond = Qi/t = ACuκCu(ΔT/LCu)

QiLCu = (ACuκCuΔT)t

QiLCu/(ACuκCuΔT) = t

t = (1,013,152.3 J)(0.181 m)/(0.0222 m2)(385 W/(m*K))(92.5 K)

t = 231.952 J/W = 231.952 (W*s)/W = 231.952 s

t [itex]\approx[/itex] 232 s

I submitted this answer to my online homework and it told me I was incorrect. I need to make sure that I am correctly analyzing the situation and accounting for everything I need to.

Any help is appreciated.

Please don't be rude. I will gladly provide more info on my calculations. Thanks.
 
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  • #2
I can think of nothing you are doing wrong, and I get the same answer.
There is one piece of information missing: we don't know the initial temperature profile of the block. But taking it to start with a linear profile with the top and bottom at 0 and 92.5C seems reasonable.
 
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  • #3
Ok, thanks! The professor told us that one of the twelve homework problems was going to give a wrong answer but he didn't tell us which so that we would still attempt all of them. This may be the problem he was talking about. I have two other problems on my homework that I haven't answered so I couldn't be sure that this was the one.
 
  • #4
So looking at the solution that the proffessor posted on this problem, the only thing that I see that he did differently is that he used kCu=390 W/(m*K) instead of kCu=385 W/(m*K).

That gives me an answer of 228.97 s
 
  • #5
castrodisastro said:
So looking at the solution that the proffessor posted on this problem, the only thing that I see that he did differently is that he used kCu=390 W/(m*K) instead of kCu=385 W/(m*K).

That gives me an answer of 228.97 s
For the temperature range in question,
http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html gives 400
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/thrcn.html gives 385
http://en.wikipedia.org/wiki/List_of_thermal_conductivities gives four estimates from 385 to 401.
So any answer in the range 223 to 232 should have been accepted.
 

1. How does thermal conductivity melt ice?

Thermal conductivity is the ability of a material to transfer heat. When a material with high thermal conductivity, such as a metal, comes into contact with ice, it quickly transfers heat to the ice, causing it to melt.

2. Can any material be used to melt ice through thermal conductivity?

Yes, any material with high thermal conductivity can be used to melt ice. However, some materials may be more effective than others. For example, metals such as aluminum and copper have very high thermal conductivity and are commonly used for melting ice.

3. How does thermal conductivity compare to other methods of melting ice?

Thermal conductivity is a very efficient way to melt ice, as it directly transfers heat to the ice without the need for additional energy. Other methods, such as using salt or hot water, also work by transferring heat to the ice, but may not be as efficient.

4. Is using thermal conductivity to melt ice safe for the environment?

Yes, using thermal conductivity to melt ice is generally considered safe for the environment. It does not involve the use of chemicals or other harmful substances, making it an environmentally friendly option.

5. Are there any limitations to using thermal conductivity to melt ice?

One limitation of using thermal conductivity to melt ice is that it requires a material with high thermal conductivity, which may not always be readily available. Additionally, it may not be as effective in melting thick layers of ice or in very cold temperatures.

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