Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu=1055J;0.3048m=1ft; and 1 °F= (5/9)°C)

Respuesta :

Answer:

223.25 [tex]$\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$[/tex]

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

[tex]$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$[/tex]

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

[tex]$1^\circ F = \frac{5}{9}^\circ C$[/tex]

We know that 1 h = 3600 s

So the thermal conductivity of the material in [tex]$\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$[/tex] is  :

Thermal conductivity :

[tex]$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$[/tex]

[tex]$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$[/tex]

=  223.25 [tex]$\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$[/tex]