Question

dos kg de. hielo a 20°C se echan en 10kg de agua a 60°C. hallar la temperstura final?

Answer 1

DATOS: Ce hielo= 2050 J/Kg.K ; Ce agua = 4180 J/Kg.K

  m1 Ce. (X – 20ºC ) = m2 Ce x (60 ºC – X)

2 Kg x 2050 J/Kg K ( x – 293 K) = 10 Kg x 4180 J/Kg.K ( 333 k – x)

4100 J/K (X – 293 K) = 41800 J/Kg ( 333 K – x)

4100 J/K x – 1201300 K = 13919400 J/K – 41800J/K x

4100 J/K + 41800J/K x = 13919400 J + 1201300  J

45900 J/K x = 15120700 J

X = 15120700 J / 45900 J/K

X = 329.42 K  = Tf

ºC 329.42 – 273 = 56.42 ºC = Tf