A very hard question about specific heat.

In summary, the problem involves finding the time it takes for a layer of ice on a pond of water at 0°C to increase from 4.50 cm to 9.00 cm thick, while the air temperature remains constant at -11.0°C. Using the heat conduction equation and the specific properties of ice, we can calculate the rate of energy extraction from the water to freeze the ice and then integrate it to find the time. The solution suggests that it takes about 42.5 seconds for the ice to increase in thickness.
  • #1
dingdong_mustaf
1
0
A pond of water at 0°C is covered with a layer of ice 4.50 cm thick. If the air temperature stays constant at -11.0°C, how much time does it take for the thickness of the ice to increase to 9.00 cm?

Hint: To solve this problem, use the heat conduction equation,

dQ/dt = kA delta T/x

and note that the incremental energy dQ extracted from the water through the thickness x is the amount required to freeze a thickness dx of ice. That is, dQ = LpA dx, where p is the density of the ice, A is the area, and L is the latent heat of fusion. (The specific gravity and thermal conductivity for ice are, respectively, 0.917 is 2.0 W/m/°C.)






I don't have much of an idea on how to attempt this question, all I've got so far is.

dQ = LpA dx
so LpA dx/dt = kA delta T/x
x/dt = L delta T/ L p dx

I guess that's useful as it gets rid of surface area in the equation ( which isn't given), but I am not sure where to go from there. Also, delta T would be zero, and so the entire equation would equal zero, which doesn't make much sense to me.

By the way I am 16 and so presume that I am very ignorant.
 
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  • #2
why should dT be 0? Outside temp is -11 while water is at 0.

dQ=KA.(T1-T2)/x .dt where T1-T2=11
dQ=mL=dx.A.P.L
so we have from above eqns.
dx.A.P.L=KA.11/x .dt
xdx.P.L=K.11 dt
integrate LHS from 0.045 to 0.09 and RHS from 0 to t, where t is the required time.
P.L.(x^2)/2=22t
substituing the values (L=3.36 x 10^5)& solving i get t=42.5 sec

IMO this is too small a value, anyway, do tell the answer :)
 
  • #3




Hi there, thank you for reaching out for help with this question. Don't worry, it's perfectly normal to feel unsure about how to approach a problem at first. I have a few tips that might help you solve this question.

First, let's define our variables and their values. We know that the thickness of the ice increases from 4.50 cm to 9.00 cm, so the change in thickness (dx) is 4.50 cm. We also know that the air temperature (delta T) is -11.0°C and the initial temperature of the water (T) is 0°C. The specific gravity (p) and thermal conductivity (k) for ice have been given, so we can use those values as well.

Now, let's look at the heat conduction equation that was provided in the hint: dQ/dt = kA delta T/x. This equation relates the rate of heat transfer (dQ/dt) to the thermal conductivity (k), surface area (A), temperature difference (delta T), and thickness (x). In this case, we are interested in finding the time it takes for the thickness of the ice (x) to increase from 4.50 cm to 9.00 cm, so we can rearrange the equation to solve for time (t):

t = x^2 / (kA delta T)

Next, we need to calculate the surface area (A) of the pond. We can assume that the pond is circular, so we can use the formula for the area of a circle (A=πr^2) to find the surface area. The radius (r) can be calculated by dividing the diameter (which is equal to the thickness of the ice) by 2. So, A = π(4.50/2)^2 = 15.90 cm^2.

Now, we have all the information we need to solve the equation for time (t). Plugging in the values, we get:

t = (4.50 cm)^2 / (2.0 W/m/°C * 15.90 cm^2 * (-11.0°C - 0°C))

t = 0.126 seconds

So, it will take approximately 0.126 seconds for the thickness of the ice to increase from 4.50 cm to 9.00 cm.

I hope this helps you understand
 

1. What is specific heat and why is it important?

Specific heat is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It is important because it helps us understand how much energy is needed to heat or cool a substance, and is crucial in many industrial processes and daily activities such as cooking.

2. How is specific heat measured?

Specific heat is measured through experiments where the amount of heat added to a substance and the resulting temperature change are recorded. The specific heat is then calculated using the equation Q = m x c x ∆T, where Q is the heat added, m is the mass of the substance, c is the specific heat, and ∆T is the change in temperature.

3. What factors affect the specific heat of a substance?

The specific heat of a substance is affected by its chemical composition, density, and phase (solid, liquid, gas). It also varies with temperature, with most substances having a higher specific heat at lower temperatures. Additionally, the presence of impurities or other substances can also affect the specific heat of a substance.

4. How does specific heat relate to thermal conductivity?

Specific heat and thermal conductivity are related, but not the same. Specific heat is a measure of a substance's ability to store heat, while thermal conductivity is a measure of its ability to transfer heat. A substance with a high specific heat will require more energy to heat up, but may not necessarily transfer that heat efficiently to other objects.

5. Can the specific heat of a substance change?

The specific heat of a substance can change with temperature, but it is generally considered to be a constant value for a given substance. However, under extreme conditions such as extreme pressures or temperatures, the specific heat of a substance may deviate from its typical value. Additionally, specific heat values may vary slightly depending on the method used to measure it.

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