A polystyrene ice chest contains 3 kg of ice at 0 degrees Celsius. Triangle H for melting = 3.34 x 10 to the 5 power J/kg p = 1000 kg/m cubed, CP =4190 J/ (kg K). Heat transfer is limited by conduction (k=0.06W/m K)) from the outside wall at 30 degrees Celsius. This means that temperature within the ice chest is not a function of position. Assume the heat flux (J/(sec m squared) is the same through all walls: top, sides, bottom. The walls have thickness = 4 cm and total surface area = 2400 cm squared.

Calculate heat flux (J/ (sec m squared) at the walls.

T within wall = 0 deg C

T outside wall = 30 deg C

wall thickness = .04 m

Heat power flux in Watts per m^2 = .06 W/m deg (30 deg)/.04 m = 45 Watts/m^2
since a Watt is a Joule per second, that is the answer.

Now the real question, which should be part B, is how long until the ice is melted :)

45 Watts/m^2 * 2400 *10^-4 m^2 = 10.8 Watts pouring out through the walls of the box

how many Joules to melt the ice?
3 kg *334000 J/kg = 10^6 Joules
so
10.8 Joules/second * T = 10^6 Joules
T = 92778 seconds to melt
or
92777/3600 = 26 hours to melt

To calculate the heat flux (J/(sec m²)) at the walls of the ice chest, we need to use the formula for one-dimensional heat conduction:

Q = (k * A * ΔT) / d

Where:
Q is the heat transfer rate (in watts)
k is the thermal conductivity of the material (in W/(m·K))
A is the cross-sectional area of heat flow (in m²)
ΔT is the temperature difference across the material (in K)
d is the thickness of the material (in meters)

First, let's convert the various measurements to the appropriate units:
Surface area (A) = 2400 cm² = 2400 * (1/100)^2 = 2.4 m²
Thickness (d) = 4 cm = 4 * (1/100) = 0.04 m
Temperature difference (ΔT) = (30 - 0) = 30 K

Now we can substitute the values into the formula to calculate the heat transfer rate (Q):

Q = (0.06 * 2.4 * 30) / 0.04 = 3.6 W

So the heat transfer rate (or heat flux) at the walls of the ice chest is 3.6 J/(sec m²).