Practica No2 quimica industrial upiicsa
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PRACTICA No. 2 Capacidad calorífica y calor de neutralización. Experimento No. 1 −m Cp ΔT = m Cp ΔT + k dT H 2 O calie nte H 2 O fría Fría Caliente V = 40ml V = 40ml T 1 = 25 ℃ T 2 = 69 ℃ Tiempo (s) 0 10 20 30 40 50 60 Temperatura ( ℃) 44 44 44 44 44 43 43 T eq = 43 ℃ m = Vδ H2O = (40 ml) (0,9778 g/ml) = 39,112g H 2 O calie nte m = Vδ H2O = (40 ml) (0,99568 g/ml) = 39,827g H 2 O fría −( 39,112 g)( 1 cal g℃ )( 43 −69 ; C) = ( 39 , 827)( 1 cal g℃ )( 43−25 ℃)+k ( 43−25 ℃) −1016,912 cal = 716,8896 cal +18 k
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Transcript of Practica No2 quimica industrial upiicsa
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
