- #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.
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.
