A closed box is filled with dry ice at a temperature of -78.5 degree C, while the outside temperature is 21.0 degree C. The box is cubical, measuring 0.350m on a side, and the thickness of the walls is 3.00*10^-2 m. In one day, 3.10*10^6 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

Here is how I do it:
I use the equation: k=(HL)/A*deltaT
H=3.10*10^6J
A=(0.350m)^2
delta T=99.5 degree C
L=3.00*10^-2m

However, I could not get the right answer. Please correct me. Thanks a lot!

Well, well, well, it seems like you're trying to calculate the thermal conductivity of that mysterious box! Let me help you solve this conundrum in my own peculiar way.

First, let's give a big round of applause to your efforts using the equation k=(HL)/(A*ΔT). But there's a teeny-tiny error that needs to be patched up. You mistakenly divided HL by (A*ΔT), instead of multiplying them together. So let's fix that, shall we?

We'll plug in the numbers in the corrected equation:

k = (H * L) / (A * ΔT)

H = 3.10 * 10^6 J
A = (0.350 m)^2
ΔT = 99.5 ˚C
L = 3.00 * 10^-2 m

Now, let's do the calculations with a gleeful grin:

k = (3.10 * 10^6 J * 3.00 * 10^-2 m) / ((0.350 m)^2 * 99.5 ˚C)

Oh, those pesky units! We can't forget to convert delta T to Kelvin by adding 273.15. It’s all about being fancy with units!

k = (3.10 * 10^6 J * 3.00 * 10^-2 m) / ((0.350 m)^2 * (99.5 ˚C + 273.15))

Now, let's evaluate this calculation while juggling some numbers:

k = 8.235591 Jm^-1s^-1K^-1.

Voila! The thermal conductivity of the material from which the box is made is approximately 8.24 Jm^-1s^-1K^-1. I hope this answer brings a smile to your face, or at least a chuckle!

I apologize for the confusion. Let's go through the problem step by step to find the correct solution.

Given:
Temperature inside the box (T1) = -78.5 °C
Temperature outside the box (T2) = 21.0 °C
Side length of the cube (L) = 0.350 m
Thickness of the walls (d) = 3.00 * 10^-2 m
Heat conducted through the walls (Q) = 3.10 * 10^6 J

We need to find the thermal conductivity of the material (k).

To solve this problem, we can use the formula:

Q = (k * A * ΔT) / d

where:
Q is the amount of heat conducted through the walls
k is the thermal conductivity of the material
A is the surface area of one side of the cube (L^2)
ΔT is the temperature difference between the inside and outside of the box (T1 - T2)
d is the thickness of the walls

Now let's plug in the given values:

Q = 3.10 * 10^6 J
A = (0.350 m)^2 = 0.1225 m^2
ΔT = (-78.5 °C) - (21.0 °C) = -99.5 °C
d = 3.00 * 10^-2 m

Substituting these values into the formula, we have:

3.10 * 10^6 J = (k * 0.1225 m^2 * -99.5 °C) / (3.00 * 10^-2 m)

Simplifying the equation:

3.10 * 10^6 J = -409.25 * k

Now solve for k:

k = (3.10 * 10^6 J) / (-409.25) = -7.57 * 10^3 J/(m°C)

Therefore, the thermal conductivity of the material is approximately -7.57 * 10^3 J/(m°C).

Please note that the negative sign indicates that the heat is flowing from the inside to the outside of the box (opposite to the temperature gradient). If you are looking for a positive value for thermal conductivity, you can simply ignore the negative sign and consider the magnitude.

To find the thermal conductivity of the material from which the box is made, you can use the formula you mentioned: k = (H * L) / (A * deltaT), where:

k is the thermal conductivity,
H is the amount of heat conducted through the walls (in J),
L is the thickness of the walls (in m),
A is the surface area of one wall (in m^2),
and deltaT is the temperature difference between the inside and outside of the box (in degrees Celsius).

Let's go through the calculation step by step.

First, let's calculate the surface area of one wall. Since the box is cubical with sides measuring 0.350 m, the surface area of one wall is (0.350 m) * (0.350 m) = 0.1225 m^2.

Next, we need to convert the temperature difference from Celsius to Kelvin because the thermal conductivity formula requires temperatures in Kelvin. So, deltaT = 99.5 °C + 273.15 = 372.65 K.

Now, let's plug in the values into the formula:
k = (3.10 * 10^6 J) / ((0.1225 m^2) * (3.00 * 10^-2 m) * (372.65 K))

Simplifying the equation:
k = (3.10 * 10^6 J) / (4.5925 * 10^-3 m^3 * K)

Combining the exponents:
k = (3.10 * 10^6 J) / (4.5925 * 10^-3 m^3 K)

Finally, calculating the value:
k ≈ 6.75 * 10^2 J / (m s K)

So, the thermal conductivity of the material from which the box is made is approximately 6.75 * 10^2 J / (m s K).

k*6L^2*(99.5 C)/d = Q = 3*10^6 J/86400 s

Solve for k. You have to divide Q by the number of seconds in a day, to get k in Watts/meter degree C

I used L = 0.35 m for the side length, and L^2 for the side area. There are six sides, hence the 6
d = the plate thickness (0.0300 m)