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.
Suppose heat transfer is due to conduction through the wall and convection at the surface with h = 8W/ (m squared K) to ambient air at T infinity = 30 degrees Celsius.
Calculate heat flux (J/ (sec m squared) at the walls.
To calculate the heat flux at the walls of the polystyrene ice chest, we first need to calculate the rate of heat transfer through conduction and convection.
1. Calculate the rate of heat transfer through conduction:
- We can use Fourier's Law of Heat Conduction, which states Q = -k * A * ΔT / d, where:
- Q is the rate of heat transfer (in watts),
- k is the thermal conductivity of the material (in watts per meter-kelvin),
- A is the surface area through which heat is transferred (in square meters),
- ΔT is the temperature difference (in kelvin), and
- d is the thickness of the material (in meters).
- In this case, we have k = 0.06 W/m-K, A = 2400 cm^2 = 0.24 m^2, ΔT = (30 - 0) K, and d = 4 cm = 0.04 m.
- Plugging in these values, we get Q_conduction = -0.06 * 0.24 * (30 - 0) / 0.04 W.
2. Calculate the rate of heat transfer through convection:
- We can use Newton's Law of Cooling, which states Q = h * A * ΔT, where:
- Q is the rate of heat transfer (in watts),
- h is the heat transfer coefficient (in watts per square meter-kelvin),
- A is the surface area through which heat is transferred (in square meters), and
- ΔT is the temperature difference (in kelvin).
- In this case, we have h = 8 W/(m^2-K), A = 0.24 m^2, and ΔT = (30 - 0) K.
- Plugging in these values, we get Q_convection = 8 * 0.24 * (30 - 0) W.
3. Calculate the total rate of heat transfer:
- Since the heat transfer through conduction and convection both occur at the same time, we can add their rates of heat transfer to get the total rate.
- Q_total = Q_conduction + Q_convection.
4. Calculate the heat flux:
- The heat flux is defined as the rate of heat transfer per unit area. So, we can calculate the heat flux by dividing the total rate of heat transfer by the surface area.
- Heat flux = Q_total / A.
Now, let's plug in the calculated values:
Q_conduction = -0.06 * 0.24 * (30 - 0) / 0.04 W
Q_convection = 8 * 0.24 * (30 - 0) W
Q_total = Q_conduction + Q_convection
Heat flux = Q_total / A
Solve these equations to find the value of the heat flux at the walls of the polystyrene ice chest.