PRACTICA No. 2Capacidad calorfica y calor de neutralizacin.

Experimento No. 1

mCpT=mCpT+kdT

H2OcalienteH2Ofra

FraCalienteV = 40mlV = 40ml T1 = 25 T2 = 69

Tiempo (s)0102030405060

Temperatura ()44444444444343

Teq = 43 m = V H2O = (40 ml) (0,9778 g/ml) = 39,112g

H2Ocaliente

m = V H2O = (40 ml) (0,99568 g/ml) = 39,827g

H2Ofra

=

=

Experimento No. 2

HClNaOHt acido = 25 t base = 24 V = 40 ml V = 40 mlTiempo (s)0102030405060

Temperatura ()31313131313131

Teq = 31

t acido + t base = ti = 24,5 m vaso = 51,1 gm vaso y solucin = 132,5gm solucin = 81,4g

NaOHac + HCl NaCl + H2O V = 40 ml V = 40 mlM = 1M M = 1Mn = 0,04 mol n = 0,04 mol

H2O = 81,4 g 2,34 g = 79,06 g Solucin 81.4g

NaCl = m = n PM = (0,04 mol)(58,5 g/mol) = 2,34 gnH2O = Cp NaCl = 4 J/g

0,04mol 614,53 cal 1mol15365,75 cal