Question

Approximately how long should it take 8.0 kg of ice at 0?C to melt when it is placed in a carefully sealed Styrofoam ice chest of dimensions 25 cm × 35 cm × 55 cm whose walls are 1.6 cm thick? Assume that the conductivity of Styrofoam is double that of air and that the outside temperature is 33 ?C.

Answer 1

We know that

Q/t = k*A*dT/dx

Q = Heat absorbed by ice to melt = m*Lf

m = mass of ice = 8.0 kg

Lf = Latent heat of fusion = 3.34*10^5 kg

k = Conductivity of styrofoam = 2*0.026 = 0.052 W/m-K

A = Area = 2*(ab + bc + ca)

A = 2*(25*35 + 35*55 + 55*25)*10^-3 m^2 = 0.835 m^2

dT = 33 - 0 = 33 C

dx = thickness = 1.6 cm = 0.016 m

Using above values:

m*Lf/t = k*A*dT/dx

t = m*Lf*dx/(k*A*dT)

t = 8*3.34*10^5*0.016/(0.052*0.835*33)

t = 29836.83 sec = 8.29 hr

Please Upvote.

Originally Conductivity of Styrofoam is = 0.023 W/m-K , it's not double of air, then we will get

t = 8*3.34*10^5*0.016/(0.023*0.835*33)

t = 67457.18 sec = 18.74 hr

So Check Both. Let me know which one works.

Please Upvote.