A Styrofoam box has a surface area of 0.75 m2 and a wall thickness of 2.0 cm. The temperature of the inner surface is 5.0°C, and the outside temperature is 24°C. If it takes 7.6 h for 5.0 kg of ice to melt in the container, determine the thermal conductivity of the Styrofoam.
To determine the thermal conductivity of the Styrofoam, we can use the rate of heat transfer equation:
Q/t = k * A * (T2 - T1) / d
Where:
Q/t is the rate of heat transfer (in watts or joules per second)
k is the thermal conductivity of the material (in watts per meter per Kelvin)
A is the surface area of the container (in square meters)
T2 is the outside temperature in Kelvin
T1 is the inside temperature in Kelvin
d is the wall thickness (in meters)
We need to rearrange the equation to solve for k:
k = (Q/t) / (A * (T2 - T1) / d)
Given that the rate of heat transfer is Q/t = 5.0 kg * 333.5 kJ/kg (latent heat of fusion of ice) / (7.6 h * 3600 s/h),
the surface area A = 0.75 m^2,
the outside temperature T2 = 24°C + 273.15 K,
the inside temperature T1 = 5°C + 273.15 K, and
the wall thickness d = 2.0 cm / 100 cm/m.
Let's plug in these values and calculate the thermal conductivity k:
k = ( (5.0 kg * 333.5 kJ/kg) / (7.6 h * 3600 s/h) ) / (0.75 m^2 * ((24°C + 273.15 K) - (5°C + 273.15 K)) / (2.0 cm / 100 cm/m) )
Simplifying the equation:
k = ( (5.0 kg * 333.5 kJ/kg) / (7.6 h * 3600 s/h) ) / (0.75 m^2 * (24 - 5)°C / (2.0 cm / 100 cm/m) )
k = ( (5.0 * 333.5) / (7.6 * 3600) ) / (0.75 * 19) m^2°C / (0.02 m)
k ≈ 0.0908 W/(m·K)
Therefore, the thermal conductivity of the Styrofoam is approximately 0.0908 W/(m·K).