transferencia de calor-balance

I. Resultados: M e = 100g H 2 O M s = 50g H 2 O Balance de Masa dm dt = 50 ∫ 2000 m dm= ∫ 0 t 50 dt m−2000 =50 t −0 dm dt = 50 Balance de Energía E e − E s = E ac E=mcpT E e − E s = dE dt m e c p T e − m s c p T e = d ( mcpT ) dt m e c p T e − m s c p T e = cp [ dm dt .T + dT dt . m ] 100 g . 4.18 J g °C .20 ° C −50 g . 4.186 J g ° C T = 4.186 [ 50 T + dT [ 50 t +2000 ] dt ] 8372−209.3 T =4.186 x 50T + 4.186 ( 50 T +2000 ) dT dt 8372− 418.6 T =( 209.3 t +8372 ) dT dt 1 209.3 [ ln| 209.3 t + 8372 | −ln| 8372 | ] = 1 −418.6 [ ln| 8372−418.6 T | − ln| 8372− 418.6 ( 62) |] −2 ln| 209.3 t +8372| +18.06 =ln| 8372−418.6 T | −9.77 e −2l n| 209.3 t +8372| +27.83 = 8372−418.6

transferencia de calor-balance

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