Calculating the efficency of a hydrogen fuel cell

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Homework Statement



I'm trying to calulate the efficency of a small hydrogen fuell cell that I built. I have all of the measurements, but I'm not coming up with a resonable effiency.

Homework Equations



[tex]\frac{P OUT}{P IN}[/tex]x 100

The power to electrolyze the water is the P IN.
3.25 W =2.5 V x .00021 Amps (.21 mA)

We measured the electrical power out finding the voltage and amperage.
P OUT = V x I
2.01 x 10^-4 W=.96 V x .00021 AMPS (.21 mA)

I caulcate the force to be 1.421 N


Then I take the electrical P Out and use it as the P IN for the car

the P IN for the car is the electrical P Out

and the P Out for the car is F x V

To find the % efficiency:

[tex]\frac{p Out of the Car}{P in Car}[/tex] x 100

The avg velocity is .0198 m/s


The Attempt at a Solution




I'm coming up with an unreasonable calculation
 
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To the Electrical power I'm coming up with:

P IN = 2.5 V x 1.3 AMPS = 3.25 W

P Out = .96 x .00021 amps = 2.01 x 10 ^ -4 W

I'm taking the P Out and using it as the power In for the car.

For the car:

P in = 2.01 x 10^-4 W
P Out = F x V
2.81 x 10^-2 J = 1.421 N x .0198 m/s

P out/ p in x 100
13980 definitely isn't reasonable.
Should my P in and P out for the car both be in Joules?

Also I'm not sure if I'm calculating the force in th right way. We used a spring scale and my measurement was .160 kg. I'm I just supposed to multiply that by 9.8 m/s^2. I know that this way gives me Newtons, but the car isn't being affected by gravity anyways. If someone could look this over I'd appreciate it.
 
can someone just let me know if I'm calculating the force the rigt way?
 
correction: it took 1.3 amps to electroylze the water