Resposta:

23)

m = massa de gelo

w = taxa de fornecimento de calor

80m = 60w

(2400 + m).5=20w

12000 + 5m = 20w

12000 + 5m = 20.4m/3 = 553,8g

w = 738, 4 cal/min

24)

a) Q1 + Q2=0 --- m1.c1.(T1-To1) + m2.c2.(T2-To2)=0 --- 1000 × 1,0 × (T - 55) + 500 × 1 × (T - 25 ) = 0 --- 1500 T = 67500 --- T = 45 °C.

b) m=1.000 + 500=1.500kg --- Q = 1500 × 1,0 × (100 - 45) = 1.500 × 55 = 82500 cal --- Q=82,5 kcal.

c) Para resfriar a água --- Q1 = - 1500 × 1 × 100 --- Q1= - 150 000 cal --- Para congelar a água --- Q2 = - 80 × 1500 cal ---

Q2=- 120 000 cal --- portanto o calor total a ser RETIRADO da água será --- Qt = Q1 + Q2 = - 270 000 cal --- Qt= - 270 kcal.

Explicação: