Termo

DESARROLLO EXAMEN PARCIAL

3)
Q= m. Ce. (Tf-28)
1000= (1Kg) (4180) (Tf-301)
1000= 4180Tf – 1258180
1259180=4180Tf
301,24 K = Tf
Q1 = Q2

(1Kg)(4180)(Te – 301,24) = (1Kg)(2000)(343 – Te)
4180Te – 1259183,2 = 686 000 – 2000Te6180Te = 1945183,2
Te = 314,75 K
Te = 314,75 – 273
Te = 41,75 C

1000 = 1Kg(130)(Tf – 323)
1000 = 130 Tf – 4199042990 = 130 Tf
330,69 = Tf

Qa = Qb
(1Kg)(130)(Te-330,69) = (1Kg)(2000)(343-Te)
130Te – 42989.70 = 686000 – 2000Te2130Te = 728989,70
Te = 342,25 – 273
Te = 68,24 C

r

5)
K = C + 273
K = 10 + 273 = 283

Qv = 2,26 x 10ª la 6 (1,6Kg) =3, 616 x 10ª la 6
Q1 = (0, 025Kg)(4180)(283 K) = 29573,5 joule

Qt = Qf + Q1 = 3,616 x 10ª la 6 + 29573,5

Qt = 3645573,5 jouleX = 3645573,5 = 729,1147 segundos
5000

6)
Qf = (0,1Kg)(3,36 x 10ª la 5) = 3,36 x 10ª la 4 joule
Q1 =(0,1Kg)(2090)(333K) = 69597 joule
Q2 = (0,1Kg)(4180)(Te – 0) = 418Te
Q3 = (0,25Kg)(4180)(40 – Te) = 327085 – 1045Te
Q4 = (5Kg)(2000)(90 –Te) = 363 x 10ª la 4 – 1 x 10ª la 4 Te

3,36 x 10 al 4 + 69597 + 418Te = 327085 – 1045Te + 363 x 10 a la 4 – 1 x 10ª la 4 Te103197 + 418Te = 3957085 – 11045Te
418Te + 11045Te = 3957085 – 103197
11463Te = 3853888
Te= 336,2 K
Te = 336,2 – 272
Te = 63,2 C