Question

Coverta: 100 kelvin em celsius, -100 fahreinheit em kelvin e 50 kelvin em Fahreinheit.

Answer 1

Relação entre as escalas

$\frac{t_{c}}{5}=\frac{t_{f}-32}{9}=\frac{t_{k}-273}{5}$

100k para C

$t_{c}=t_{k}-273\\t_{c}=100-273\\t_{c}=-173\degree C$

-100F para k

$t_{k}=\frac{5t_{f}+2297}{9}$

$t_{k}=\frac{5.-100+2297}{9}$

$t_{k}=\frac{-500+2297}{9}$

$t_{k}=\frac{1797}{9}$

$t_{k}=199,66k$

50k em F

$t_{f}=\frac{9t_{k}-2297}{5}$

$t_{f}=\frac{9.50-2297}{5}$

$t_{f}=\frac{450-2297}{5}$

$t_{f}=-73,88\degree F$