Answer:
a).a. 5K
b. 5°C
b). 480 W
Step-by-step explanation:
Given:
Thickness of mortar, L = 20 cm = 0.2 m
Inside temperature,
= 21°C
=21+273 = 294 K
Outside temperature,
= 26°C
=26+273 = 299 K
a). Temperature difference in --
a) Kelvin
ΔT =
-

= 299 - 294
= 5 K
b). Celcius
ΔT =
-

= 26 -21
= 5°C
b). We know thermal conductivity for cement mortar is k = 1.2 W / m-K
For steady state we know


W