Part A.- In the winter, a pond has a layer of ice on top that is becoming thicke
ID: 1422393 • Letter: P
Question
Part A.- In the winter, a pond has a layer of ice on top that is becoming thicker. The air above the layer is at a constant -11 C and the water below is at a constant 0 C. Starting from zero thickness, how much time is needed for the layer of ice to reach a thickness of 25 cm ? (Hint: Call the thickness through which the heat conducts x, then express dQ in the heat conduction equation in terms of dx, then separate t and x and integrate both sides.)
Part B.- Assuming that the depth of the pond is 50 m at all points, calculate how much time would be needed for the pond to freeze entirely.
Explanation / Answer
let thickness of ice deposited=x m
then heat transferred through an additional layer of thickness dx
is given by dQ/dt=k*A*(0-(-11))/x
where k=thermal conductivity of ice
A=cross sectional area of the ice layer
dQ/dt=k*A*11/x...(1)
to turn the dx layer of water into ice, heat required
=mass*latent heat of fusion
=density of water*volume*latent heat of fusion
=1000*A*dx*334*10^3
=334*10^6*A*dx
hence using dQ=334*10^6*A*dx
equation 1 becomes:
334*10^6*A*dx/dt=k*A*11/x
using k=2.2 W/(m.K)
==>334*10^6*dx/dt=2.2*11/x
==>13.8*10^6*x*dx=dt
integrating both sides,
13.8*10^6*0.5*x^2=t
using limits of x from x=0 to x=0.25 cm,
time taken to freeze the ice of 25 cm thickness
=13.8*10^6*0.5*0.25^2=4.313*10^5 seconds
=119.8 hours
part B:
using limits of x =0 to 50 , we get time taken to freeze the entire pond
=t=13.8*10^6*0.5*50^2=1.725*10^10 seconds