40 Reamur =.. Rn 112 F=...K 473 K=...Rn 80R= ....K Mohon di bantu kakak Jawaban ni

112F = (112 + 456, 67) 5/ 9
= 317,59k

1. )
40°R = 40°R

2. )
112°F = 317,4 K

T°F 》T°C
T°C = 5/9 × ( T°F - 32 )
T°C = 5/9 × ( 112 - 32 )
T°C = 5/9 × 80
T°C = 400/9
T°C = 44,4444 ...
T°C = 44,4

T°C 》TK
TK = T°C + 273
TK = 44,4 + 273
TK = 317,4

3. )
473 K = 160 R

TK 》T°C
T°C = TK - 273
T°C = 473 - 273
T°C = 200

T°C 》T°R
T°R = 4/5 × T°C
T°R = 4/5 × 200
T°R = 4 × 40
T°R = 160

4. )
80 R = 373 K

T°R 》T°C
T°C = 5/4 × T°R
T°C = 5/4 × 80
T°C = 5 × 20
T°C = 100

T°C 》TK
TK = T°C + 273
TK = 100 + 273
TK = 373

-----------
Catatan !
T°C ⇒suhu dalam derajat Celcius
T°R ⇒suhu dalam derajar Reamur
T°F ⇒suhu dalam derajat Fahrenheit
TK ⇒suhu dalam derajat Kelvin
-------------------------------

Semoga Bermanfaat !