Sam, Elijah, Joy Those are 100% correct from my human knowledge
Answer:
its so simple. u must mind some formulas.
Explanation:
mili->10^(-3)
micro->10^(-6)
nano->10^(-9)
so write the exact number and move "." to left or right depend question.
in this one:
23.5 is 23500000.0 nano with a default dot at the end
for turning to mili u must move the dot 6 steps to left so it will be: 23.5 milimeter.
Answer:
N_A=1.5*10^-8 kmol/s.m^2
Explanation:
<u>KNOWN: </u>
Molar concentration of helium at the inner and outer surfaces of a plastic membrane. Diffusion coefficient and membrane thickness.
<u>FIND:</u>
Molar diffusion flux.
<u>ASSUMPTIONS:</u>
(1) Steady-state conditions, (2) One-dimensional diffusion in a plane wall, (3) Stationary medium, (4) Uniform C = C_A + C_B.
<u>ANALYSIS:</u> The molar flux may be obtained from
N_A=D_AB/L(C_A,1-C_A,2)
=10^-9 m^2/s/0.001 m(0.02-0.005)kmol/m^3
N_A=1.5*10^-8 kmol/s.m^2
<u>COMMENTS:</u> The mass flux is:
n_A,x=M_a*N_A,x
n_A,x=6*10^-8 kg/s m^2
Answer:
178 kJ
Explanation:
Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:
ΔU = -Q = -c*m*ΔT (1)
where c= specific heat of water = 4180 J/kg*ºC
m= total mass = 6*0.355 Kg = 2.13 kg
ΔT = difference between final and initial temperatures = 20ºC
Replacing by these values in (1), we can solve for Q as follows:
Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ
So, the amount of heat transfer from the six canned drinks is 178 kJ.
Answer:
The detailed answer to the question is explained in the attached file.
Explanation:
