Question

A closed box is filled with dry ice at a temperature of -89.8 oC, while the outside temperature is 18.3 oC. The box is cubical, measuring 0.276 m on a side, and the thickness of the walls is 3.08 × 10-2 m. In one day, 3.75 × 106 J of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

Answer 1

The thermal conductivity of the material from which the box is made is
`2.34*10^(3)\ W?m^2`.

Step-by-step explanation:

Given that,

Inside temperature = -89.8°C = 183.2 K

Outside temperature = 18.3 °C = 291 K

Thickness of walls
`x=3.08*10^(-2)`

Heat
`Q= 3.75*10^(6)\ J`

Side = 0.276 m

We need to calculate the thermal conductivity of the material

Using formula of the thermal conductivity

`k=(Q\Delta x)/(A(T_(2)-T_(1)))`

Put the value into the formula

`k=(3.75*10^(6)*3.08*10^(-2))/(6*(0.276^2)(291-183.2))`

`k=2344.195\ W/m^2`

`k=2.34*10^(3)\ W?m^2`

Hence, The thermal conductivity of the material from which the box is made is
`2.34*10^(3)\ W?m^2`.