Answer:
a).a. 5K
b. 5°C
b). 480 W
Explanation:
Given:
Thickness of mortar, L = 20 cm = 0.2 m
Inside temperature, [tex]T_{in}[/tex] = 21°C
=21+273 = 294 K
Outside temperature, [tex]T_{out}[/tex] = 26°C
=26+273 = 299 K
a). Temperature difference in --
a) Kelvin
ΔT = [tex]T_{out}[/tex] - [tex]T_{in}[/tex]
= 299 - 294
= 5 K
b). Celcius
ΔT = [tex]T_{out}[/tex] - [tex]T_{in}[/tex]
= 26 -21
= 5°C
b). We know thermal conductivity for cement mortar is k = 1.2 W / m-K
For steady state we know
[tex]Q = \frac{\Delta T}{\frac{L}{K.A}}[/tex]
[tex]Q = \frac{5}{\frac{0.2}{1.2\times 2\times 2}}[/tex]
[tex]Q = 480[/tex] W