Question

The temperature of a system drops by 45°F during a cooling process. Express this drop in temperature in K, R, and °C

Answer 1

Answer:

280.3722 K,504.67°R ,7.222°C

Explanation:

FAHRENHEIT TO KELVIN: T(K)=(T(F°)+459.67)$\frac{5}{9}$

we have to convert 45°F=(45+459.67)×$\frac{5}{9}$

=280.3722

FAHRENHEIT TO RANKINE :T(R°)=T(F°)+459.67

we have to convert 45°F=45+459.67

=504.67

FAHRENHEIT TO CELSIUS:T(C°)=$\frac{F-32}{9}$×5

we have to convert 45°F=$\frac{45-32}{9}$×5

=7.22°C