Resposta :
Q1 = m1 x c1 x ∆T
Q1 = 200.1.(Tf - 20)
Q1 = 200Tf - 4000
Q2 = m2 x c2 x ∆T
Q2 = 80.0,4.(Tf - 100)
Q = 32Tf - 3200
Q1 + Q2 = 0
200Tf - 4000 + 32Tf - 3200 = 0
232Tf = 7200
Tf = 7200/232
Tf ≈ 31,03 °C
Q1 = m1 x c1 x ∆T
Q1 = 200.1.(Tf - 20)
Q1 = 200Tf - 4000
Q2 = m2 x c2 x ∆T
Q2 = 80.0,4.(Tf - 100)
Q = 32Tf - 3200
Q1 + Q2 = 0
200Tf - 4000 + 32Tf - 3200 = 0
232Tf = 7200
Tf = 7200/232
Tf ≈ 31,03 °C