Conversio scalarum temperaturae (scala Celsiana, scala Fahrenheit, scala Kelviniana)

9 Conversiones scalarum temperaturarum (scala Celsiana, scala Fahrenheit, scala Kelviniana)

1. 50 ^oC = … ^oF?

Solutio

In atmosphaera normali Curabitur, punctum congelationis aquae est 0 ^oC in Scala Celsiana et 32 ^oF in scala Fahrenheit. Sub pressione atmosphaerica normali, punctum ebullitionis aquae est 100. ^oC in scala Celsii et 212 ^oF in scala Fahrenheit.

0 ^o32 C = ^oF et 100 ^o212 C = ^oF. Mutatio quinque graduum Celsii^o = mutatio 9°F^o.

Pro scala Celsiana, distantia inter 0 ^oC et LV ^oC in 100 intervalla aequalia divisa. Pro scala Fahrenheit, distantia inter 0 ^oC et LV ^oC in 180 intervalla aequalia divisa.

T^oF = (180/100) T^oC + 32

T^oF = (9/5) T^oC + 32

T^oF = (9/5) 50 + 32

T^oF = (9) 10 + 32

T^oF = 90 + 32

T^oF = 122

50 ^o122 C = ^oF

2. 86 ^oF = … ^oC?

Solutio

T^oC = (100/180)(T)^oF – 32)

T^oC = (5/9)(T)^oF – 32)

T^oC = (5/9)(86 – 32)

T^oC = (5/9)(54)

T^oC = (5)(6)

T^o30 C =

86 ^oF = 30 ^oC

3. 50^oC = ... K?

Solutio

T = T ^oC + 273

T = 20 000 + 10 000

XI T =

50 ^oC K 323

4. 212^oF = ... K?

Solutio

T^oC = (100/180)(T)^oF – 32)

T^oC = (5/9)(T)^oF – 32)

T^oC = (5/9)(212 – 32)

T^oC = (5/9)(180)

T^oC = (5)(20)

T^o100 C =

212 ^oF = 100 ^oC + 273

212 ^oF = 373 K

 

5. x ^oC = x ^oF

x = ... ?

Solutio

1: Scala Celsii in scalam Fahrenheit convertenda

2: Scala Fahrenheit in scalam Celsii convertens

6. 122°F = ... Celsius

Solutio

Conversio inter duas scalas temperaturae scribi potest:

T_C = 5/9 (T)_F - 32)

T_C = Calor in Celsiis, T_F = temperatura in Fahrenheit

Temperatura in gradibus Celsii:

T_C = 5/9 (122 – 32) = T_C = 5/9 (90) = 5 (10)

T_C = 50 ^oC

7. Figura infra ostendit mensura temperaturae a Liquor cum thermometro scalae Fahrenheit! Si temperatura liquidi thermometro scalae Celsius metitur, tum quid est temperatura liquidie.

Notum:

Fahrenheit IX scale (T_F) II =^oF

Quaesitum: Scala Celsiana

solution:

Ad pressionem unius atmosphaerae, punctum congelationis aquae is 0°C cum scala Fahrenheit 32 sit. ^oF. Contra, tpunctum ebullitionis aquae pro CElsius scala est centum ^oC dum scala Fahrenheit is 212 ^oF.

In scala Celsius, inter 0°C et 100°C est 100°, dum in scala Fahrenheit inter 32°F et 212°F est 180°.

T_C = 100/180 (T)_F - 32)

T_C = 5/9 (T)_F - 32)

T_C = 5/9 (95 - 32)

T_C = 5/9 (63)

T_C 315/9 =

T_C = 35^oC

8. Ex figura infra, determina t.Temperatura P in thermometro Celsiano.

Solutio

T_C = 100/180 (T)_F - 32)

T_C = 5/9 (T)_F - 32)

T_C = 5/9 (104 – 32)

T_C = 5/9 (72)

T_C 360/9 =

T_C = 40 ^oC

9. Si temperatura scalae Celsius est ut in figura infra ostenditur, temperaturam scalae Fahrenheit determina ut in figura infra demonstratur.

solution:

T^oF = (180/100) T^oC + 32

T^oF = (9/5) T^oC + 32

T^oF = (9/5) 60 + 32

T^oF = (9) 12 + 32

T^oF = 108 + 32

T^oF = 140

1. Scalas temperaturae convertens
2. Dilatatio linearis
3. Expansio areae
4. Expansio voluminis
5. Calor
6. Aequivalens mechanicum caloris
7. Calor specificus et capacitas calorica
8. Calor latens, calor fusionis, calor vaporisationis
9. Conservatio energiae ad translationem caloris