Răspuns :
Q=m*c*delta t
m = Q/c * delta t
m = 2090/ 4,18 * (70 - 20) = 2090 / (4,18 * 50) = 10 kg apa
Q = nr moli * delta H
nr moli = Q/delta H = 312890/3128,9 = 100 moli
masa molara C6H6 = 6*12+6 = 78 g/mol
=> masa benzen = 100 * 78 = 7800 g benzen
m = Q/c * delta t
m = 2090/ 4,18 * (70 - 20) = 2090 / (4,18 * 50) = 10 kg apa
Q = nr moli * delta H
nr moli = Q/delta H = 312890/3128,9 = 100 moli
masa molara C6H6 = 6*12+6 = 78 g/mol
=> masa benzen = 100 * 78 = 7800 g benzen
1. Q=mc(t2 - t1)⇒ m= Q /cΔt = 2090kj /4180kj/kg·grd·50grd = 0,01kg =100g
2. La arderea unui mol de benzen se degajă 3128,9 kj Q= m×ΔHcombustie ⇒ m=Q/ΔH = 312890 /3128,9 = 100 moli benzen m=nr.moli×M =100×78 =7800g = 7,8 kg