Question

1_ transforme as temperaturas abaixo : A) 104 °F em °K. B) -114,1 °C em °F C) 78,5 °C em °K D) 305 °K em °F e em °C. E) 25 °C em °F e em °K​

Answer 1

Resposta:

A) 313°K

B) 237,38°F

C) 351,5°K

D) -49,5°F/-45,25°C

E) 77°F/318°K

Explicação:

A) TF=104

TK=?

Tk-273 = tf-32

5 9

Tk-273 = 104-32

5 9

Tk-273 = 72

5 9

72÷9= 8

Tk-273 = 8

5

8×5 =40

Tk-273= 40

Tk=40+273

Tk= 313°K

B) Tc= -114,1

Tf=?

Tf-32 = tc

9 5

Tf-32 = -114,1

9 5

-114,1÷5= 22,82

Tf-32 = 22,82

9

22,82×9= 205,38

Tf-32=205,38

Tf=205,38+32

Tf=237,38°F

C) Tk=?

Tc=78,5

Tk-273 = 78,5

5 5

Corta os 5

Tk-273= 78,5

Tk= 78,5+273

Tk= 351,5 °K

D) Tk=305

Tf=?

Tk-273 = tf-32

5 9

305-273 = tf-32

9 5

32 = tf-32

9 5

32÷9=3,5

3,5 = tf-32

5

3,5×5 =17,5

17,5 =tf-32

Tf= -32-17,5

Tf= -49,5°F

Tf= -49,5

Tc=?

Tf-32 = Tc

9 5

-49,5-32 = Tc

9 5

-81,5 = Tc

9 5

-81,5÷9= 9,05

-9,05 = Tc

5

-9,05×5=-45,25

Tc=-45,25 °C

E) Tc=25

Tf=?

Tf-32 = 25

9 5

Tf-32 = 5

9

9×5= 45

Tf-32= 45

Tf=45+32

Tf= 77°F

Tf= 77

Tk=?

Tk-273 = tf-32

5 9

Tk-273 = 77-32

5 9

Tk-273 = 45

5 9

45÷9=5

Tk-273 = 5

9

9×5= 45

Tk= 45+273= 318

Tk=318°K