Calculate how long it will take for the ice to melt?

  • Thread starter Thread starter klilly
  • Start date Start date
  • Tags Tags
    Ice
klilly
Messages
3
Reaction score
0
Calculate how long it will take for the ice to melt??

Homework Statement



A chilly bin has walls 5.90 cm thick and the total area of the walls is 0.700 m2. The chilly bin is loaded with 2.00 kg of ice at 0.00 °C and stood on a rack so that its entire surface is in contact with the air. The temperature on the outside of the chilly bin is 28.0 °C. If the chilly bin is made of styrofoam (kstyrofoam = 0.0100 J s–1 m–1 °C–1), how many hours will it take to melt all of the ice?
(Note: Lfwater = 3.35 × 105 J kg–1)


Homework Equations



I was given (L (water) x m (water))/ kA(Change in T/Change in thickness)
But it didnt work out right


The Attempt at a Solution



Answer is: 56 hours
I have no idea how to get to this :(
 
Physics news on Phys.org


law of thermal conduction says

\mathcal{P}=kA\;\frac{\Delta T}{\Delta x}

where P is the power transferred, k is thermal conductivity, A is the area of the surface through which energy will flow, \frac{\Delta T}{\Delta x} is temperature gradient.
\Delta T is the temperature difference between inner and outer surface,
\Delta x is the thickness of the bin.

now we have been given , for chilly bin (made out of styrofoam)

k= 0.0100 J s-1 m-1 °C-1
A=0.700 m2
delta T=28-0=28 °C
delta x=5.92 x 10-2 m

using this you can find the power which transfers from the outside to inside where ice is
stored. P= 3.322 J s-1=3.322 W

now amount of energy required to melt m kg of ice is

Q=m_{ice}L_f

mice=2 kg ; Lf=3.35 × 105 J kg-1

so we get Q= 6.7 x 105 J

if t is the time required to melt all ice then Q must be equal to P x t . solve for t. its 56 hrs
 
  • Like
Likes Mbywater94
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Back
Top