Atividade 6 Física 34 -wps Office

Mariana Elvira de Silva\nATIVIDADE 06\nFÍSICA II\nCALORIMETRIA\n\n1. 400 mL DE CAFÉ\n c = cm\n Q = cmΔt\n m:At\n g: 50°C\n GARDAFA 30% NO EQUILÍBRIO: QCAFÉ + QGARDAFA = 0\n EQUILÍBRIO A 80°C ASSIM:\n a: 10 cal/g°C\n cmst + cmst = 0\n d: 1g/mL\n 1.400(80-0) + cm(80-30) = 0\n -400 + cm 50 = 0\n -400 + cm 0 = 0\n cm = 80 cal/°C\n\nC: c = cm\n c = 80 cal/°C\n\n2. 1 kg = 1000 g\n 30. K5*x\n x = 20000 J\n c = Q\n c = mΔt\n (B)\n\n3. m = 2 - 80°C\n 420g - 20°C\n EQUILÍBRIO A 70°C\n cágua 1,4x10³ J/kg°C\n m = 180000\n 42000\n m = 450 g\n (B) Mariana Elvira G. Silva\n4. m = 600g\n t: 10°C\n Q1 = mcΔt\n Q1 = 600 · 0,50 · 10 = 3,000 cal\n a = 25000 cal\n Lf = 20 cal/g\n c = 0,50 cal/g°C\n mF = 9\n (D)\n\n5. 1 Kg = 1000 g\n -20°C - 100°C\n Q1 = mcΔt\n Q1 = 1000 · 0,95 · 20 = 10,000 cal\n c = 1 cal/g°C\n cágua 1 cal/g°C\n Lf = 80 cal/g\n\nQ2 = m·Lf\n Q2 = 1000 · 80 = 80,000 cal\n Q3 = mcΔt\n Q3 = 1000 · 1 · 100,000 cal\n Q4 = m·Lv\n Q4 = m · 100,000 + 540,000 cal\n Q2 = 10,000 + 80,000 + 100,000 + 540,000\n Q2 = 730,000 cal , 730 Kcal\n\n6. 300 cal/min\n ϕ = Q\n 30 = Q/30\n z = TEMPO\n (D)\n C = Q = 300 - 30 cal/°C\n ΔT = 50-20 Mariana Elvira G. Silva\n7. m = 400g\n ϕ = Q = 600 cal/min = Q\n Δt\n T1 = 22°C\n T2 = 72°C\n t = 5 min\n ϕ = 600 cal/min\n c: Q = 3000 cal\n c = Q/mΔT\n 400(72-22) \n c = 0,15 cal/g°C\n\nb) Isolamento térmico, ou seja, dificulta as trocas de calor por condução de meio exterior para corpo da pessoa e vice-versa.\n\n2. 300 mL, 80°C\n mcΔt + mcΔt = 0\n 300·1·(Tg-80) + 200·1·(Tg-10) = 0\n c = 1,0 cal/g°C\n 500·Tg + 26000 = 0\n 500Tg = 26000\n Tg = 52°C\n\n10. 50 ml café 60°C\n 50 · 1 (50-90) + m · 1 (50-30) = 0\n -2000 + m20 = 0\n m = 2000/20\n m = 100 mL