Question

PROBLEM 6.2 The thermal conductivity of a material varies with temperature as $ln k = 0.01T + 0.5$ where $k$ has the units $W cdot m^{-1} cdot K^{-1}$ and $T$ has the units degrees Celsius. Heat flows by conduction through a plane slab of this material of thickness 0.1 m, the left face of which is at 100°C and the right face of which is at 0°C. Calculate: (a) The mean thermal conductivity of the material in the range 0 to 100°C (b) The heat flux through the slab (c) The actual variation of $T$ with distance through the slab (d) The temperature at $x = 0.05$ m (e) The temperature at $x = 0.05$ m calculated using the mean thermal conduc- tivity (f) The actual temperature gradient in the slab at $x = 0.05$ m (g) The temperature gradient at $x = 0.05$ m calculated using the mean thermal conductivity

          PROBLEM 6.2
The thermal conductivity of a material varies with temperature as
$ln k = 0.01T + 0.5$
where $k$ has the units $W cdot m^{-1} cdot K^{-1}$ and $T$ has the units degrees Celsius. Heat flows
by conduction through a plane slab of this material of thickness 0.1 m, the left face
of which is at 100°C and the right face of which is at 0°C. Calculate:
(a) The mean thermal conductivity of the material in the range 0 to 100°C
(b) The heat flux through the slab
(c) The actual variation of $T$ with distance through the slab
(d) The temperature at $x = 0.05$ m
(e) The temperature at $x = 0.05$ m calculated using the mean thermal conduc-
tivity
(f) The actual temperature gradient in the slab at $x = 0.05$ m
(g) The temperature gradient at $x = 0.05$ m calculated using the mean thermal
conductivity

PROBLEM 6.2
The thermal conductivity of a material varies with temperature as
ln k = 0.01T + 0.5
where k has the units W cdot m^-1 cdot K^-1 and T has the units degrees Celsius. Heat flows
by conduction through a plane slab of this material of thickness 0.1 m, the left face
of which is at 100°C and the right face of which is at 0°C. Calculate:
(a) The mean thermal conductivity of the material in the range 0 to 100°C
(b) The heat flux through the slab
(c) The actual variation of T with distance through the slab
(d) The temperature at x = 0.05 m
(e) The temperature at x = 0.05 m calculated using the mean thermal conduc-
tivity
(f) The actual temperature gradient in the slab at x = 0.05 m
(g) The temperature gradient at x = 0.05 m calculated using the mean thermal
conductivity

Added by Sergio H.

Question

Please give Ace some feedback