Resposta :
Q1 + Q2 + Q3 = 0
mc(Tf-Ti) - C(Tf-Ti) + mc(Tf-Ti) = 0
100 x 1 x (40-20) - C x (40-20) + 20 x 0,5 x (40-100)= 0
2000 - 20C - 600= 0
-20C + 1400 = 0
-20C = -1400
C = 1400/20
C = 70 cal/°C
mc(Tf-Ti) - C(Tf-Ti) + mc(Tf-Ti) = 0
100 x 1 x (40-20) - C x (40-20) + 20 x 0,5 x (40-100)= 0
2000 - 20C - 600= 0
-20C + 1400 = 0
-20C = -1400
C = 1400/20
C = 70 cal/°C