The approach Oz suggested is absolutely fine. Just some extra comments on the formulas:
Originally Posted by Adeeb  I know those formulas (not including in questions):
P = I 4 (pi) r^2. |
This is the equation to get the irradiance (per unit area), P, at a certain distance, r, away from a point source with a total irradiance, I. That equation isn't going to be helpful here because you don't have a point source of radiation in your problem; you're already given the irradiance per unit area.
Note that $\displaystyle 4 \pi r^2$ is the surface area of a sphere.
for ice melt: dQ/dt = d/dt (m x Lf) = d/dt [ (p=density) x V x lf)
(Lf = 334 KJ/kg)
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This is an equation that's calculating the rate of change of latent heat transfer and substituting the mass with the equation valid for a homogeneous medium, which is very close to what you need.
Do you remember the definition of power? It is
$\displaystyle P = \frac{dE}{dt}$
which is the rate of change of energy transfer with time. If the power is constant over some interval $\displaystyle \Delta t$, then
$\displaystyle P = \frac{\Delta E}{\Delta t}$
and you can rearrange this to get
$\displaystyle \Delta t = \frac{\Delta E}{P}$
This is basically the idea that if you divide a total "thing" by it's constant "thing per second", you get the time taken to get the total "thing" in seconds, but applied to energy. It's no different from the solution to this: "If I buy beer at a constant rate of 2 pints per hour and I drink 8 pints, how many hours did it take?". You divide the total (8 pints) by the rate (2 pints/hour) to get the time (4 hours).
You can do the same thing for your problem question. You should consider the total latent heat transfer ($\displaystyle \Delta Q = \Delta m l = \rho \Delta V l$) and because there's no variation in the irradiance over time, you can just consider a constant heat transfer rate
$\displaystyle \dot{Q} = \frac{\Delta Q}{ \Delta t}$
(i.e. over one second) and then, once you've found the total amount of heat required to melt the ice, you just divide that total heat by the heat rate to get the time required.
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