No idea , sorry. maybe it is.....actually i dont know, sooo sorry
Answer:
efficiency of heating with this oven is 51 %
Explanation:
to raise the temp of 200 ml of coffee from 30°C to 60°C the energy input to microwave oven is:
1100 J/s x 45 = 49,500 J
AT 100% efficiency
For 1°C the energy required to raise the temperature of 1 ml = 4.2 J
So for 30 C°, 1°C the energy required to raise the temperature of 200 ml =
Q = (4.2) (200)(30) = 25,200 J
efficiency = 25,200/49,500 = 0.51 = 51%
Answer:
-490.7 K
Explanation:
Given:
[Ni^2+]= 0.4 M
[Pb^2+]=0.002 M
∆V= -0.012 V
VNi= -0.250V
VPb= -0.126V
F= 96500 C
R= 8.314 JK-1 mol-1
n= 2
From
T= -nF/R [∆V-(VNi-VPb)/ln [Pb2+]/[Ni2+]]
T= 2(96500)/8.314[ (-0.012) -(-0.250) - (-0.126))/ln[0.002]/[0.4]
T= 23213.856(0.112/(-5.298))
T= -490.7 K
Answer:
the most best answer is 50.1 you never find such type of answer
