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.

Question

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 the outside surface temperature.

Answer 1

To calculate the outside surface temperature of the polystyrene ice chest, we need to consider the heat transfer through conduction and convection.

First, let's calculate the heat transfer through conduction.

1. Calculate the thermal resistance of the wall:
The thermal resistance (R) is given by the formula R = (thickness)/(conductivity x area).
- The thickness of the wall is 4 cm, which is equal to 0.04 m.
- The conductivity (k) is given as 0.06 W/(m K).
- The total surface area of the walls is given as 2400 cm^2, which is equal to 0.24 m^2.

So, R = (0.04 m) / (0.06 W/(m K) x 0.24 m^2) = 0.2778 K/W.

2. Calculate the heat flux through the wall:
The heat flux (q) is given by the formula q = (difference in temperature) / (thermal resistance).
- The difference in temperature is the outside temperature (T_outside) minus the ice temperature (0 degrees Celsius = 273 K).

Let's denote the outside surface temperature as T_outside.

We can use the heat flux equation with the given information to solve for T_outside:
q = (T_outside - 273 K) / 0.2778 K/W.

Now, let's consider the heat transfer through convection.

3. Calculate the heat transfer coefficient (h):
The heat transfer coefficient (h) is given as 8 W/(m^2 K).

4. Calculate the convective heat transfer rate:
The convective heat transfer rate (Q_conv) is given by the formula Q_conv = h x area x (T_outside - T_infinity).
- The area is the total surface area of the walls, which is 0.24 m^2.
- T_infinity is the ambient air temperature, given as 30 degrees Celsius = 303 K.

Using the convective heat transfer equation with the given information, we can solve for T_outside:
Q_conv = (8 W/(m^2 K)) x 0.24 m^2 x (T_outside - 303 K).

Since the heat transfer is the same through all walls, we can equate the heat flux through conduction (q) and the convective heat transfer rate (Q_conv).

(T_outside - 273 K) / 0.2778 K/W = (8 W/(m^2 K)) x 0.24 m^2 x (T_outside - 303 K).

To solve this equation, we can rearrange it and solve for T_outside.

Please note that the equation involves an iterative calculation method, which requires multiple iterations to obtain a accurate solution.

Therefore, the outside surface temperature, T_outside, can be determined by solving the above equation using an iterative calculation method or by using software capable of solving non-linear equations.