1- Em julho de 1988 a temperatura atingiu –10 oC no Rio Grande do Sul. (a) Calcule a que taxa perde calor uma parede de tijolo de 6 m × 4 m e 13 cm de

 (a) ∆Q/∆t = 4.5 × 103 W,. (b) 51.5 kW, (c) 200 W

só usar a formula  (∆Q/∆t) = (k* A *∆T)/d

A variação de temperatura é de 33°C  pois:

-10° a 23° = 33°

Portanto:

a)

\(\[\begin{align} & {{P}_{cond}}=\frac{\Delta Q}{\Delta t}=\frac{KA\Delta T}{L} \\ & {{P}_{cond}}=\frac{0,74.24.33}{0,13} \\ & {{P}_{cond}}=4.508\text{x1}{{\text{0}}^{3}}\text{W} \\ & {{P}_{cond}}=\text{4}\text{.5 KW} \\ \end{align}\] \)

b)

Considerando uma condutividade térmica do vidro = 0,8:

\(\[\begin{align} & {{P}_{cond}}=\frac{\Delta Q}{\Delta t}=\frac{KA\Delta T}{L} \\ & {{P}_{cond}}=\frac{4,68.0,8.33}{0,003} \\ & {{P}_{cond}}=41.184\text{x1}{{\text{0}}^{3}}\text{W} \\ & {{P}_{cond}}=\text{41}\text{.18 KW} \\ \end{align}\] \)

c)

\(\[\begin{align} & {{\text{P}}_{\text{cond}}}\text{=}\frac{\text{ }\!\!\Delta\!\!\text{ Q}}{\text{ }\!\!\Delta\!\!\text{ t}}\text{=}\frac{\text{KA }\!\!\Delta\!\!\text{ T}}{\text{L}} \\ & {{\text{P}}_{\text{cond}}}\text{=}\frac{\text{4}\text{.68}\text{.0}\text{,8}\text{.0}\text{,026}}{\text{0}\text{,026}} \\ & {{\text{P}}_{\text{cond}}}\text{=154W} \\ \end{align}\] \)