A Styrofoam box has a surface area of 0.78 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 Styrofoam, we can use the formula for heat transfer through a material:

Q/t = (k*A*(T2 - T1))/d

where:
Q/t is the rate of heat transfer (in this case, it's the energy required to melt the ice in the Styrofoam box divided by the time),
k is the thermal conductivity of Styrofoam,
A is the surface area of the Styrofoam box,
T2 is the outside temperature,
T1 is the inside temperature, and
d is the wall thickness of the Styrofoam box.

First, we need to calculate the energy required to melt the ice in the Styrofoam box. We can use the formula Q = m*L, where:
Q is the energy required to melt the ice,
m is the mass of the ice, and
L is the latent heat of fusion for ice (which is 334,000 J/kg).

Given:
m (mass of ice) = 5.0 kg
L (latent heat of fusion for ice) = 334,000 J/kg

Q = m * L
Q = 5.0 kg * 334,000 J/kg
Q = 1,670,000 J

Next, we can substitute the known values into the heat transfer formula:

1,670,000 J / (7.6 h * 3600 s/h) = (k * 0.78 m^2 * (24°C - 5.0°C)) / 0.02 m

Simplifying, we have:

k = (1,670,000 J / (7.6 h * 3600 s/h)) * (0.02 m) / (0.78 m^2 * (24°C - 5.0°C))

k = (1,670,000 J / 27480 s) * (0.02 m) / (0.78 m^2 * 19°C)

Solving this equation will give us the thermal conductivity of Styrofoam.

To determine the thermal conductivity of the Styrofoam, we can use the equation:

Q = (k * A * ΔT) / d

Where:
Q is the heat transferred (in Joules)
k is the thermal conductivity of the material (in W/m*K)
A is the surface area of the Styrofoam box (in m^2)
ΔT is the temperature difference between the inner and outer surfaces (in K)
d is the thickness of the wall (in meters)

First, we need to calculate the heat transferred, Q. By using the equation:

Q = m * L

Where:
m is the mass of ice melted (in kg)
L is the latent heat of fusion for ice, which is 334,000 J/kg

Given the mass of ice (m) is 5.0 kg, we can calculate Q:

Q = 5.0 kg * 334,000 J/kg = 1,670,000 J

Now, we can rearrange the formula to solve for thermal conductivity, k:

k = (Q * d) / (A * ΔT)

Given:
A = 0.78 m^2
d = 2.0 cm = 0.02 m
ΔT = 24°C - 5.0°C = 19 K

Plugging in the values:

k = (1,670,000 J * 0.02 m) / (0.78 m^2 * 19 K)

Simplifying:

k = 42,080,000 J / (14.82 m^2 * K)

k ≈ 2,838,330 J / (m^2 * K)

Therefore, the thermal conductivity of the Styrofoam is approximately 2,838,330 J / (m^2 * K).