
#1
Sep1911, 08:30 PM

P: 3

1. The problem statement, all variables and given/known data
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) 2. Relevant equations I was given (L (water) x m (water))/ kA(Change in T/Change in thickness) But it didnt work out right 3. The attempt at a solution Answer is: 56 hours I have no idea how to get to this :( 



#2
Sep1911, 09:21 PM

P: 607

law of thermal conduction says
[tex]\mathcal{P}=kA\;\frac{\Delta T}{\Delta x}[/tex] where P is the power transferred, k is thermal conductivity, A is the area of the surface through which energy will flow, [itex]\frac{\Delta T}{\Delta x}[/itex] is temperature gradient. [itex]\Delta T[/itex] is the temperature difference between inner and outer surface, [itex]\Delta x[/itex] 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 m^{2} delta T=280=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 [tex]Q=m_{ice}L_f[/tex] m_{ice}=2 kg ; L_{f}=3.35 × 105 J kg^{1} so we get Q= 6.7 x 10^{5} 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 


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