A closed box is filled with dry ice at a temperature of -81.5 C, while the outside temperature is 27.4 C. The box is cubical, measuring 0.307 m on a side, and the thickness of the walls is 4.47 × 10-2 m. In one day, 3.37 × 106 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.
To find the thermal conductivity of the material from which the box is made, we need to use the formula for heat conduction:
Q = (k * A * (T1 - T2)) / d
where:
Q = amount of heat conducted (in joules)
k = thermal conductivity of the material (in watts per meter per degree Celsius)
A = total surface area of the box (in square meters)
T1 = temperature on one side of the box (in degrees Celsius)
T2 = temperature on the other side of the box (in degrees Celsius)
d = thickness of the box wall (in meters)
In this case, we are given:
Q = 3.37 × 10^6 J (amount of heat conducted)
A = 6 * (0.307 m)^2 (total surface area of the box)
T1 = 27.4°C (outside temperature)
T2 = -81.5°C (temperature inside the box)
d = 4.47 × 10^-2 m (thickness of the box wall)
Now, we can plug in these values into the formula and solve for k:
3.37 × 10^6 J = (k * 6 * (0.307 m)^2 * (27.4°C - (-81.5°C))) / (4.47 × 10^-2 m)
Simplifying, we have:
3.37 × 10^6 J = (k * 6 * (0.307 m)^2 * 108.9°C) / (4.47 × 10^-2 m)
To isolate k, we can rearrange the equation:
k = (3.37 × 10^6 J * 4.47 × 10^-2 m) / (6 * (0.307 m)^2 * 108.9°C)
Calculating this expression will give us the thermal conductivity of the material from which the box is made.