How Do You Calculate Ice Thickness in a Thermal Conductivity Problem?

[Paged]

Homework Statement

So, we have ice at a steady state, and above it is Air at -5.2 celcius, and below it is water at 3.98 C. The total depth of the ice and the water is 1.42m, and they want us to find the thickness of the ice. The thermal conductivies (k) of ice and water are 1.67 and 0.502 respectively.

Homework Equations

Rate of heat flow=(k)(A)(delta T)/(l)

The Attempt at a Solution

It says the ice is in a steady state, so there is an even flow of heat to the air and to the water. So I know I'm setting two equations equal too each other. I also know the area is not given, but it will cancel in the two equated equations. What I don't know is what delta T i am meant to be measuring, and where the depth of the ice fits in at all. Any point in the right direction would be appriciated

 

[anathema]

Hello, it seems like you have all the necessary information to solve this problem. The first thing you should do is to write out the two equations for the rate of heat flow in the ice and in the water. The equation you provided is correct, but just to clarify, the two equations should be:

Rate of heat flow in ice = (k_ice)(A)(delta T_ice)/l_ice

Rate of heat flow in water = (k_water)(A)(delta T_water)/l_water

Where k is the thermal conductivity, A is the area, delta T is the temperature difference, and l is the thickness of the material.

Since the ice is in a steady state, the rate of heat flow in the ice must be equal to the rate of heat flow in the water. This means you can set the two equations equal to each other:

(k_ice)(A)(delta T_ice)/l_ice = (k_water)(A)(delta T_water)/l_water

As you mentioned, the area will cancel out, so you can ignore it for now. Now, you just need to plug in the values you have for the thermal conductivities, temperatures, and thicknesses. The only unknown in this equation is the thickness of the ice, which you can solve for.

To do this, you will need to use the fact that the total depth of the ice and water is 1.42m. This means that the thickness of the ice and the thickness of the water must add up to 1.42m. So, you can write the following equation:

l_ice + l_water = 1.42m

You now have two equations with two unknowns (l_ice and l_water). You can solve this system of equations to find the thickness of the ice.

I hope this helps guide you in the right direction. Let me know if you have any further questions.

 

[m2003]

I would approach this problem by first identifying the variables and their relationships. The rate of heat flow is directly proportional to the thermal conductivity (k) and the temperature difference (delta T), and inversely proportional to the length (l) over which the heat is flowing. In this case, we are interested in finding the thickness of the ice, which is represented by the variable l.

To solve this problem, we can use the equation for rate of heat flow and set it equal to each other for the ice and water layers, as you have correctly identified. We can also assume that the temperature at the interface between the ice and water is constant, as it is in a steady state.

Using this information, we can set up the following equations:

Rate of heat flow through ice = Rate of heat flow through water

(kice)(A)(delta Tice)/(lice) = (kwater)(A)(delta Twater)/(lwater)

We can then rearrange this equation to solve for the thickness of the ice, lice:

lice = (kwater)(delta Twater)/(kice)(delta Tice) * lwater

Plugging in the given values for k and delta T, we can calculate the thickness of the ice to be approximately 0.92 meters.

It is important to note that the depth of the ice (1.42m) is not directly used in this calculation. Instead, it is indirectly accounted for through the thermal conductivities and temperature differences.

In conclusion, by setting up and equating the equations for rate of heat flow through the ice and water layers, we can solve for the thickness of the ice and determine that it is approximately 0.92 meters.