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

Answer:

The thermal conductivity of the material from which the box is made is $2.34\times10^{3}\ W?m^2$.

Explanation:

Given that,

Inside temperature = -89.8°C = 183.2 K

Outside temperature = 18.3 °C = 291 K

Thickness of walls $x=3.08\times10^{-2}$

Heat $Q= 3.75\times10^{6}\ J$

Side = 0.276 m

We need to calculate the thermal conductivity of the material

Using formula of the thermal conductivity

$k=\dfrac{Q\Delta x}{A(T_{2}-T_{1})}$

Put the value into the formula

$k=\dfrac{3.75\times10^{6}\times3.08\times10^{-2}}{6\times(0.276^2)(291-183.2)}$

$k=2344.195\ W/m^2$

$k=2.34\times10^{3}\ W?m^2$

Hence, The thermal conductivity of the material from which the box is made is $2.34\times10^{3}\ W?m^2$.