Question

1) t ( °C ) = 0
t ( K ) = ?

2) t ( ° C ) = 10
t ( ° F ) = ?

3) ∆t = 5 °C
∆T = ?

Answer 1

1) t ( K ) = t °C ) + 273,15
t ( K ) = 0 + 273,15
t = 273,15 K.

2) t ( °F ) = 9/5 • t ( °C ) + 32
t ( °F ) = 9/5 • 10 + 32
t ( °F ) = 18 + 32
t = 50 °F.

3) ∆T=∆t
∆t = 5 °C
∆T = 5 K.


Sper că te-am ajutat !

Answer 2

t(K)=273,15+t(C)

t(K)=273,15+0=273,15K


t(F)=9t(C)/5+32
t(F)=9*10/5 +32=18+32=50F

∆T=T1-T2=(t1+273,15)-(t2+273,15)=t1-t2=>∆T=∆t=5K(C)